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Trigonometric Functions

Class 11 · Maths · Chapter 3 · Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions with fully worked steps for radians, quadrant values, identity proofs and half-angle values

As per current NCERT edition.

Miscellaneous Exercise

Q1. Prove that: 2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π13=02\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}=0

Sol.

LHS=2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π13 \begin{array}{l} \displaystyle \text{LHS}=2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13} \end{array}

Using cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},

=2cos⁡π13cos⁡9π13+2cos⁡3π13+5π132cos⁡3π13−5π132 \begin{array}{l} \displaystyle =2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+2\cos\frac{\dfrac{3\pi}{13}+\dfrac{5\pi}{13}}{2}\cos\frac{\dfrac{3\pi}{13}-\dfrac{5\pi}{13}}{2} \end{array}
=2cos⁡π13cos⁡9π13+2cos⁡4π13cos⁡−π13 \begin{array}{l} \displaystyle =2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+2\cos\frac{4\pi}{13}\cos\frac{-\pi}{13} \end{array}

Using cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta,

=2cos⁡π13cos⁡9π13+2cos⁡4π13cos⁡π13 \begin{array}{l} \displaystyle =2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+2\cos\frac{4\pi}{13}\cos\frac{\pi}{13} \end{array}
=2cos⁡π13(cos⁡9π13+cos⁡4π13) \begin{array}{l} \displaystyle =2\cos\frac{\pi}{13}\left(\cos\frac{9\pi}{13}+\cos\frac{4\pi}{13}\right) \end{array}

Using cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},

=2cos⁡π13⋅2cos⁡9π13+4π132cos⁡9π13−4π132 \begin{array}{l} \displaystyle =2\cos\frac{\pi}{13}\cdot2\cos\frac{\dfrac{9\pi}{13}+\dfrac{4\pi}{13}}{2}\cos\frac{\dfrac{9\pi}{13}-\dfrac{4\pi}{13}}{2} \end{array}
=2cos⁡π13⋅2cos⁡13π13×2cos⁡5π13×2=2cos⁡π13⋅2cos⁡π2cos⁡5π26 \begin{array}{l} \displaystyle =2\cos\frac{\pi}{13}\cdot2\cos\frac{13\pi}{13\times2}\cos\frac{5\pi}{13\times2} \\[1em] \displaystyle =2\cos\frac{\pi}{13}\cdot2\cos\frac{\pi}{2}\cos\frac{5\pi}{26} \end{array}

Using cos⁡π2=0\cos\frac{\pi}{2}=0,

=2cos⁡π13⋅2×0⋅cos⁡5π26 \begin{array}{l} \displaystyle =2\cos\frac{\pi}{13}\cdot2\times0\cdot\cos\frac{5\pi}{26} \end{array}
=0=RHS \begin{array}{l} \hspace{0.8em} \displaystyle =0=\text{RHS} \end{array}

Hence proved.

Q2. Prove that: (sin⁡3x+sin⁡x)sin⁡x+(cos⁡3x−cos⁡x)cos⁡x=0(\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x=0

Sol.

LHS=(sin⁡3x+sin⁡x)sin⁡x+(cos⁡3x−cos⁡x)cos⁡x=sin⁡3xsin⁡x+sin⁡2x+cos⁡xcos⁡3x−cos⁡2x=cos⁡3xcos⁡x+sin⁡3xsin⁡x−cos⁡2x+sin⁡2x \begin{array}{l} \displaystyle \text{LHS}=(\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x \\[0.4em] \hspace{0.8em} \displaystyle =\sin3x\sin x+\sin^2x+\cos x\cos3x-\cos^2x \\[0.4em] \hspace{0.8em} \displaystyle =\cos3x\cos x+\sin3x\sin x-\cos^2x+\sin^2x \end{array}

Using cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B)=\cos A\cos B+\sin A\sin B,

=cos⁡(3x−x)−(cos⁡2x−sin⁡2x) \begin{array}{l} \hspace{0.8em} \displaystyle =\cos(3x-x)-(\cos^2x-\sin^2x) \end{array}

Using cos⁡2x−sin⁡2x=cos⁡2x\cos^2x-\sin^2x=\cos2x,

=cos⁡2x−cos⁡2x \begin{array}{l} \hspace{0.8em} \displaystyle =\cos2x-\cos2x \end{array}
=0=RHS \begin{array}{l} \hspace{0.8em} \displaystyle =0=\text{RHS} \end{array}

Hence proved.

Q3. Prove that: (cos⁡x+cos⁡y)2+(sin⁡x−sin⁡y)2=4cos⁡2x+y2(\cos x+\cos y)^2+(\sin x-\sin y)^2=4\cos^2\frac{x+y}{2}

Sol.

LHS=(cos⁡x+cos⁡y)2+(sin⁡x−sin⁡y)2=cos⁡2x+cos⁡2y+2cos⁡xcos⁡y+sin⁡2x+sin⁡2y−2sin⁡xsin⁡y \begin{array}{l} \displaystyle \text{LHS}=(\cos x+\cos y)^2+(\sin x-\sin y)^2 \\[0.4em] \hspace{0.8em} \displaystyle =\cos^2x+\cos^2y+2\cos x\cos y+\sin^2x+\sin^2y-2\sin x\sin y \end{array}

Using sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1,

=1+1+2(cos⁡xcos⁡y−sin⁡xsin⁡y) \begin{array}{l} \hspace{0.8em} \displaystyle =1+1+2(\cos x\cos y-\sin x\sin y) \end{array}

Using cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos(x+y)=\cos x\cos y-\sin x\sin y,

=2+2cos⁡(x+y) \begin{array}{l} \hspace{0.8em} \displaystyle =2+2\cos(x+y) \end{array}
=2(1+cos⁡(x+y)) \begin{array}{l} \hspace{0.8em} \displaystyle =2(1+\cos(x+y)) \end{array}

Using cos⁡2θ=2cos⁡2θ−1\cos2\theta=2\cos^2\theta-1,

=2(1+2cos⁡2x+y2−1) \begin{array}{l} \hspace{0.8em} \displaystyle =2\left(1+2\cos^2\frac{x+y}{2}-1\right) \end{array}
=4cos⁡2x+y2=RHS \begin{array}{l} \hspace{0.8em} \displaystyle =4\cos^2\frac{x+y}{2}=\text{RHS} \end{array}

Hence proved.

Q4. Prove that: (cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2=4sin⁡2x−y2(\cos x-\cos y)^2+(\sin x-\sin y)^2=4\sin^2\frac{x-y}{2}

Sol.

LHS=(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2=cos⁡2x+cos⁡2y−2cos⁡xcos⁡y+sin⁡2x+sin⁡2y−2sin⁡xsin⁡y \begin{array}{l} \displaystyle \text{LHS}=(\cos x-\cos y)^2+(\sin x-\sin y)^2 \\[0.4em] \displaystyle =\cos^2x+\cos^2y-2\cos x\cos y+\sin^2x+\sin^2y-2\sin x\sin y \end{array}

Using sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1,

=1+1−2(cos⁡xcos⁡y+sin⁡xsin⁡y) \begin{array}{l} \displaystyle =1+1-2(\cos x\cos y+\sin x\sin y) \end{array}

Using cos⁡(x−y)=cos⁡xcos⁡y+sin⁡xsin⁡y\cos(x-y)=\cos x\cos y+\sin x\sin y,

=2−2cos⁡(x−y) \begin{array}{l} \hspace{0.8em} \displaystyle =2-2\cos(x-y) \end{array}
=2(1−cos⁡(x−y)) \begin{array}{l} \displaystyle =2(1-\cos(x-y)) \end{array}

Using cos⁡2θ=1−2sin⁡2θ\cos2\theta=1-2\sin^2\theta,

=2(1−(1−2sin⁡2x−y2)) \begin{array}{l} \displaystyle =2\left(1-\left(1-2\sin^2\frac{x-y}{2}\right)\right) \end{array}
=2(1−1+2sin⁡2x−y2)\displaystyle =2\left(1-1+2\sin^2\frac{x-y}{2}\right)
=4sin⁡2x−y2=RHS\displaystyle =4\sin^2\frac{x-y}{2}=\text{RHS}

Hence proved.

Q5. Prove that:

sin⁡x+sin⁡3x+sin⁡5x+sin⁡7x=4cos⁡xcos⁡2xsin⁡4x\textcolor{#C34E36}{ \begin{array}{l} \displaystyle \sin x+\sin3x+\sin5x+\sin7x \\[0.4em] \hspace{1.8em} \displaystyle =4\cos x\cos2x\sin4x \end{array} }

Sol.

LHS=sin⁡x+sin⁡3x+sin⁡5x+sin⁡7x \begin{array}{l} \displaystyle \text{LHS}=\sin x+\sin3x+\sin5x+\sin7x \end{array}

Using sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},

=2sin⁡x+3x2cos⁡x−3x2+2sin⁡5x+7x2cos⁡5x−7x2 \begin{array}{l} \displaystyle =2\sin\frac{x+3x}{2}\cos\frac{x-3x}{2}+2\sin\frac{5x+7x}{2}\cos\frac{5x-7x}{2} \end{array}
=2sin⁡2xcos⁡(−x)+2sin⁡6xcos⁡(−x) \begin{array}{l} \displaystyle =2\sin2x\cos(-x)+2\sin6x\cos(-x) \end{array}

Using cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta,

=2sin⁡2xcos⁡x+2sin⁡6xcos⁡x \begin{array}{l} \displaystyle =2\sin2x\cos x+2\sin6x\cos x \end{array}
=2cos⁡x(sin⁡2x+sin⁡6x) \begin{array}{l} \displaystyle =2\cos x(\sin2x+\sin6x) \end{array}

Using sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},

=2cos⁡x(2sin⁡2x+6x2cos⁡2x−6x2) \begin{array}{l} \displaystyle =2\cos x\left(2\sin\frac{2x+6x}{2}\cos\frac{2x-6x}{2}\right) \end{array}
=2cos⁡x(2sin⁡4xcos⁡(−2x)) \begin{array}{l} \displaystyle =2\cos x(2\sin4x\cos(-2x)) \end{array}

Using cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta,

=4cos⁡xsin⁡4xcos⁡2x=RHS \begin{array}{l} \displaystyle =4\cos x\sin4x\cos2x=\text{RHS} \end{array}

Hence proved.

Q6. Prove that: (sin⁡7x+sin⁡5x)+(sin⁡9x+sin⁡3x)(cos⁡7x+cos⁡5x)+(cos⁡9x+cos⁡3x)=tan⁡6x\frac{(\sin7x+\sin5x)+(\sin9x+\sin3x)}{(\cos7x+\cos5x)+(\cos9x+\cos3x)}=\tan6x

Sol.

LHS=(sin⁡7x+sin⁡5x)+(sin⁡9x+sin⁡3x)(cos⁡7x+cos⁡5x)+(cos⁡9x+cos⁡3x) \begin{array}{l} \displaystyle \text{LHS}=\frac{(\sin7x+\sin5x)+(\sin9x+\sin3x)}{(\cos7x+\cos5x)+(\cos9x+\cos3x)} \end{array}

Using sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},
cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},

=2sin⁡7x+5x2cos⁡7x−5x2+2sin⁡9x+3x2cos⁡9x−3x22cos⁡7x+5x2cos⁡7x−5x2+2cos⁡9x+3x2cos⁡9x−3x2 \begin{array}{l} \hspace{0.4em} \displaystyle =\frac{2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2}+2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2}}{2\cos\frac{7x+5x}{2}\cos\frac{7x-5x}{2}+2\cos\frac{9x+3x}{2}\cos\frac{9x-3x}{2}} \end{array}
=2sin⁡6xcos⁡x+2sin⁡6xcos⁡3x2cos⁡6xcos⁡x+2cos⁡6xcos⁡3x=2sin⁡6x(cos⁡x+cos⁡3x)2cos⁡6x(cos⁡x+cos⁡3x)=tan⁡6x=RHS \begin{array}{l} \hspace{0.4em} \displaystyle =\frac{2\sin6x\cos x+2\sin6x\cos3x}{2\cos6x\cos x+2\cos6x\cos3x} \\[1em] \hspace{0.4em} \displaystyle =\frac{2\sin6x(\cos x+\cos3x)}{2\cos6x(\cos x+\cos3x)} \\[0.8em] \hspace{0.4em} \displaystyle =\tan6x=\text{RHS} \end{array}

Hence proved.

Q7. Prove that: sin⁡3x+sin⁡2x−sin⁡x=4sin⁡xcos⁡x2cos⁡3x2\sin3x+\sin2x-\sin x=4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}

Sol.

LHS=sin⁡3x+sin⁡2x−sin⁡x=(sin⁡3x−sin⁡x)+sin⁡2x \begin{array}{l} \displaystyle \text{LHS}=\sin3x+\sin2x-\sin x \\[0.4em] \hspace{0.8em} \displaystyle =(\sin3x-\sin x)+\sin2x \end{array}

Using sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},

=2cos⁡3x+x2sin⁡3x−x2+sin⁡2x \begin{array}{l} \hspace{0.8em} \displaystyle =2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2}+\sin2x \end{array}
=2cos⁡2xsin⁡x+sin⁡2x \begin{array}{l} \hspace{0.8em} \displaystyle =2\cos2x\sin x+\sin2x \end{array}

Using sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x,

=2cos⁡2xsin⁡x+2sin⁡xcos⁡x \begin{array}{l} \hspace{0.8em} \displaystyle =2\cos2x\sin x+2\sin x\cos x \end{array}
=2sin⁡x(cos⁡2x+cos⁡x) \begin{array}{l} \hspace{0.8em} \displaystyle =2\sin x(\cos2x+\cos x) \end{array}

Using cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},

=2sin⁡x(2cos⁡2x+x2cos⁡2x−x2) \begin{array}{l} \hspace{0.8em} \displaystyle =2\sin x\left(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2}\right) \end{array}
=4sin⁡xcos⁡x2cos⁡3x2=RHS \begin{array}{l} \hspace{0.8em} \displaystyle =4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}=\text{RHS} \end{array}

Hence proved.

Q8. Find sin⁡x2\sin\frac{x}{2}, cos⁡x2\cos\frac{x}{2} and tan⁡x2\tan\frac{x}{2}: tan⁡x=−43\tan x=-\frac43, xx in quadrant II

Sol. Since xx lies in quadrant II,

π2<x<ππ4<x2<π2 \begin{array}{l} \displaystyle \frac{\pi}{2}<x<\pi \\[0.8em] \displaystyle \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2} \end{array}

So, x2\frac{x}{2} lies in quadrant I, and sin⁡x2\sin\frac{x}{2}, cos⁡x2\cos\frac{x}{2}, tan⁡x2\tan\frac{x}{2} are all positive.

We know that

sec⁡2x=1+tan⁡2x\textcolor{#1B5E20}{ \sec^2x=1+\tan^2x }

Substituting the value of tan⁡x\tan x,

sec⁡2x=1+(−43)2=1+169=259 \sec^2x=1+\left(-\frac43\right)^2=1+\frac{16}{9}=\frac{25}{9}

Since xx lies in quadrant II, sec⁡x\sec x is negative, so sec⁡x=−53\sec x=-\frac53 and cos⁡x=−35\cos x=-\frac35.

Using 2sin⁡2x2=1−cos⁡x2\sin^2\frac{x}{2}=1-\cos x,

2sin⁡2x2=1−(−35)=85\displaystyle 2\sin^2\frac{x}{2}=1-\left(-\frac35\right)=\frac85
sin⁡2x2=45\displaystyle \sin^2\frac{x}{2}=\frac45
sin⁡x2=25=255\displaystyle \sin\frac{x}{2}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}

Using 2cos⁡2x2=1+cos⁡x2\cos^2\frac{x}{2}=1+\cos x,

2cos⁡2x2=1+(−35)=1−35=25\displaystyle 2\cos^2\frac{x}{2}=1+\left(-\frac35\right)=1-\frac35=\frac25
cos⁡2x2=15\displaystyle \cos^2\frac{x}{2}=\frac15
cos⁡x2=15=55\displaystyle \cos\frac{x}{2}=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}

Therefore,

tan⁡x2=sin⁡x2cos⁡x2=2515=2 \tan\frac{x}{2}=\frac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}=\frac{\dfrac{2}{\sqrt5}}{\dfrac{1}{\sqrt5}}=2

∴ sin⁡x2=255\sin\frac{x}{2}=\frac{2\sqrt5}{5}, cos⁡x2=55\cos\frac{x}{2}=\frac{\sqrt5}{5} and tan⁡x2=2\tan\frac{x}{2}=2.

Q9. Find sin⁡x2\sin\frac{x}{2}, cos⁡x2\cos\frac{x}{2} and tan⁡x2\tan\frac{x}{2}: cos⁡x=−13\cos x=-\frac13, xx in quadrant III

Sol. Since xx lies in quadrant III,

π<x<3π2π2<x2<3π4 \begin{array}{l} \displaystyle \pi<x<\frac{3\pi}{2} \\[0.8em] \displaystyle \frac{\pi}{2}<\frac{x}{2}<\frac{3\pi}{4} \end{array}

So, x2\frac{x}{2} lies in quadrant II; sin⁡x2\sin\frac{x}{2} is positive, while cos⁡x2\cos\frac{x}{2} and tan⁡x2\tan\frac{x}{2} are negative.

Using 2sin⁡2x2=1−cos⁡x2\sin^2\frac{x}{2}=1-\cos x,

2sin⁡2x2=1−(−13)=43\displaystyle 2\sin^2\frac{x}{2}=1-\left(-\frac13\right)=\frac43
sin⁡2x2=23\displaystyle \sin^2\frac{x}{2}=\frac23
sin⁡x2=23=63\displaystyle \sin\frac{x}{2}=\sqrt{\frac23}=\frac{\sqrt6}{3}

Using 2cos⁡2x2=1+cos⁡x2\cos^2\frac{x}{2}=1+\cos x,

2cos⁡2x2=1+(−13)=1−13=23\displaystyle 2\cos^2\frac{x}{2}=1+\left(-\frac13\right)=1-\frac13=\frac23
cos⁡2x2=13\displaystyle \cos^2\frac{x}{2}=\frac13
cos⁡x2=−13=−33\displaystyle \cos\frac{x}{2}=-\frac{1}{\sqrt3}=-\frac{\sqrt3}{3}

Therefore,

tan⁡x2=sin⁡x2cos⁡x2=63−33=−2 \tan\frac{x}{2}=\frac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}=\frac{\dfrac{\sqrt6}{3}}{-\dfrac{\sqrt3}{3}}=-\sqrt2

∴ sin⁡x2=63\sin\frac{x}{2}=\frac{\sqrt6}{3}, cos⁡x2=−33\cos\frac{x}{2}=-\frac{\sqrt3}{3} and tan⁡x2=−2\tan\frac{x}{2}=-\sqrt2.

Q10. Find sin⁡x2\sin\frac{x}{2}, cos⁡x2\cos\frac{x}{2} and tan⁡x2\tan\frac{x}{2}: sin⁡x=14\sin x=\frac14, xx in quadrant II

Sol. Since xx lies in quadrant II,

π2<x<ππ4<x2<π2 \begin{array}{l} \displaystyle \frac{\pi}{2}<x<\pi \\[0.8em] \displaystyle \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2} \end{array}

So, x2\frac{x}{2} lies in quadrant I, and sin⁡x2\sin\frac{x}{2}, cos⁡x2\cos\frac{x}{2}, tan⁡x2\tan\frac{x}{2} are all positive.

We know that

cos⁡2x=1−sin⁡2x\textcolor{#1B5E20}{ \cos^2x=1-\sin^2x }

Substituting the value of sin⁡x\sin x,

cos⁡2x=1−(14)2=1−116=1516 \cos^2x=1-\left(\frac14\right)^2=1-\frac{1}{16}=\frac{15}{16}

Since xx lies in quadrant II, cos⁡x\cos x is negative, so cos⁡x=−154\cos x=-\frac{\sqrt{15}}{4}.

Using 2sin⁡2x2=1−cos⁡x2\sin^2\frac{x}{2}=1-\cos x,

2sin⁡2x2=1−(−154)=1+154=4+154\displaystyle 2\sin^2\frac{x}{2}=1-\left(-\frac{\sqrt{15}}{4}\right)=1+\frac{\sqrt{15}}{4}=\frac{4+\sqrt{15}}{4}
sin⁡2x2=4+158=8+21516\displaystyle \sin^2\frac{x}{2}=\frac{4+\sqrt{15}}{8}=\frac{8+2\sqrt{15}}{16}
sin⁡x2=8+2154\displaystyle \sin\frac{x}{2}=\frac{\sqrt{8+2\sqrt{15}}}{4}

Using 2cos⁡2x2=1+cos⁡x2\cos^2\frac{x}{2}=1+\cos x,

2cos⁡2x2=1+(−154)=1−154=4−154\displaystyle 2\cos^2\frac{x}{2}=1+\left(-\frac{\sqrt{15}}{4}\right)=1-\frac{\sqrt{15}}{4}=\frac{4-\sqrt{15}}{4}
cos⁡2x2=4−158=8−21516\displaystyle \cos^2\frac{x}{2}=\frac{4-\sqrt{15}}{8}=\frac{8-2\sqrt{15}}{16}
cos⁡x2=8−2154\displaystyle \cos\frac{x}{2}=\frac{\sqrt{8-2\sqrt{15}}}{4}

Therefore,

tan⁡x2=sin⁡x2cos⁡x2=8+2158−215 \tan\frac{x}{2}=\frac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}=\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8-2\sqrt{15}}}

Multiplying the numerator and the denominator by 8+215\sqrt{8+2\sqrt{15}},

=8+215(8−215)(8+215)\displaystyle =\frac{8+2\sqrt{15}}{\sqrt{(8-2\sqrt{15})(8+2\sqrt{15})}}
=8+21564−60\displaystyle =\frac{8+2\sqrt{15}}{\sqrt{64-60}}
=8+2152\displaystyle =\frac{8+2\sqrt{15}}{2}
=4+15\displaystyle =4+\sqrt{15}

∴ sin⁡x2=8+2154\sin\frac{x}{2}=\frac{\sqrt{8+2\sqrt{15}}}{4}, cos⁡x2=8−2154\cos\frac{x}{2}=\frac{\sqrt{8-2\sqrt{15}}}{4} and tan⁡x2=4+15\tan\frac{x}{2}=4+\sqrt{15}.

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