Class 11 Maths NCERT Solutions Chapter 3 Miscellaneous Exercise

Using the product-to-sum formula, 2\cos A\cos B=\cos(A+B)+\cos(A-B), on the first term with A = π/13, B = 9π/13:

2\cos\dfrac{\pi}{13}\cos\dfrac{9\pi}{13}=\cos\dfrac{10\pi}{13}+\cos\left(-\dfrac{8\pi}{13}\right)=\cos\dfrac{10\pi}{13}+\cos\dfrac{8\pi}{13}

So the LHS becomes:

\cos\dfrac{10\pi}{13}+\cos\dfrac{8\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{5\pi}{13}

Now, using \cos(\pi-\theta)=-\cos\theta:

\cos\dfrac{10\pi}{13}=\cos\left(\pi-\dfrac{3\pi}{13}\right)=-\cos\dfrac{3\pi}{13}

\cos\dfrac{8\pi}{13}=\cos\left(\pi-\dfrac{5\pi}{13}\right)=-\cos\dfrac{5\pi}{13}

Substituting back:

-\cos\dfrac{3\pi}{13}-\cos\dfrac{5\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{5\pi}{13}=0

Using the sum-to-product formulas:

\sin3x+\sin x=2\sin2x\cos x

\cos3x-\cos x=-2\sin2x\sin x

Substituting into the LHS:

\text{LHS}=(2\sin2x\cos x)\sin x+(-2\sin2x\sin x)\cos x

=2\sin2x\sin x\cos x-2\sin2x\sin x\cos x

Expand both squared terms:

\text{LHS}=\cos^2x+2\cos x\cos y+\cos^2y+\sin^2x-2\sin x\sin y+\sin^2y

Group using \sin^2\theta+\cos^2\theta=1 for x and for y:

=(\cos^2x+\sin^2x)+(\cos^2y+\sin^2y)+2(\cos x\cos y-\sin x\sin y)

=1+1+2\cos(x+y)=2+2\cos(x+y)=2(1+\cos(x+y))

Using the half-angle identity 1+\cos\theta=2\cos^2\dfrac{\theta}{2} with θ = x+y:

=2\times2\cos^2\dfrac{x+y}{2}=4\cos^2\dfrac{x+y}{2}

Expand both squared terms:

\text{LHS}=\cos^2x-2\cos x\cos y+\cos^2y+\sin^2x-2\sin x\sin y+\sin^2y

Group using \sin^2\theta+\cos^2\theta=1:

=1+1-2(\cos x\cos y+\sin x\sin y)=2-2\cos(x-y)=2(1-\cos(x-y))

Using the half-angle identity 1-\cos\theta=2\sin^2\dfrac{\theta}{2} with θ = x−y:

=2\times2\sin^2\dfrac{x-y}{2}=4\sin^2\dfrac{x-y}{2}

Group the outer pair and the inner pair:

\text{LHS}=(\sin x+\sin7x)+(\sin3x+\sin5x)

Apply the sum-to-product formula to each pair:

\sin x+\sin7x=2\sin4x\cos3x

\sin3x+\sin5x=2\sin4x\cos x

So:

\text{LHS}=2\sin4x\cos3x+2\sin4x\cos x=2\sin4x(\cos3x+\cos x)

Apply the sum-to-product formula again to (cos 3x + cos x):

\cos3x+\cos x=2\cos2x\cos x

So:

\text{LHS}=2\sin4x\times2\cos2x\cos x=4\cos x\cos2x\sin4x

Simplify the numerator using the sum-to-product formula:

\sin7x+\sin5x=2\sin6x\cos x

\sin9x+\sin3x=2\sin6x\cos3x

Numerator = 2\sin6x\cos x+2\sin6x\cos3x=2\sin6x(\cos x+\cos3x)

Simplify the denominator the same way:

\cos7x+\cos5x=2\cos6x\cos x

\cos9x+\cos3x=2\cos6x\cos3x

Denominator = 2\cos6x\cos x+2\cos6x\cos3x=2\cos6x(\cos x+\cos3x)

Dividing:

\dfrac{2\sin6x(\cos x+\cos3x)}{2\cos6x(\cos x+\cos3x)}=\dfrac{\sin6x}{\cos6x}=\tan6x

Group so that a sine difference appears:

\text{LHS}=(\sin3x-\sin x)+\sin2x

Apply the sum-to-product formula to (sin 3x − sin x):

\sin3x-\sin x=2\cos2x\sin x

Also write sin 2x = 2 sin x cos x:

\text{LHS}=2\cos2x\sin x+2\sin x\cos x=2\sin x(\cos2x+\cos x)

Apply the sum-to-product formula to (cos 2x + cos x):

\cos2x+\cos x=2\cos\dfrac{3x}{2}\cos\dfrac{x}{2}

So:

\text{LHS}=2\sin x\times2\cos\dfrac{3x}{2}\cos\dfrac{x}{2}=4\sin x\cos\dfrac{x}{2}\cos\dfrac{3x}{2}

Since x lies in quadrant II, π/2 < x < π, so π/4 < x/2 < π/2 — x/2 lies in quadrant I, where sin, cos and tan are all positive.

\sec^2x=1+\tan^2x=1+\dfrac{16}{9}=\dfrac{25}{9}

In quadrant II, cos x is negative, so sec x is negative: sec x = −5/3

2\sin^2\dfrac{x}{2}=1-\cos x=1-\left(-\dfrac{3}{5}\right)=\dfrac{8}{5} \;\Rightarrow\; \sin^2\dfrac{x}{2}=\dfrac{4}{5}

2\cos^2\dfrac{x}{2}=1+\cos x=1+\left(-\dfrac{3}{5}\right)=\dfrac{2}{5} \;\Rightarrow\; \cos^2\dfrac{x}{2}=\dfrac{1}{5}

Both are positive in quadrant I, so:

\tan\dfrac{x}{2}=\dfrac{\sin(x/2)}{\cos(x/2)}=\dfrac{2/\sqrt5}{1/\sqrt5}=2

Since x lies in quadrant III, π < x < 3π/2, so π/2 < x/2 < 3π/4 — x/2 lies in quadrant II, where sin is positive but cos and tan are negative.

2\sin^2\dfrac{x}{2}=1-\cos x=1-\left(-\dfrac{1}{3}\right)=\dfrac{4}{3} \;\Rightarrow\; \sin^2\dfrac{x}{2}=\dfrac{2}{3}

2\cos^2\dfrac{x}{2}=1+\cos x=1+\left(-\dfrac{1}{3}\right)=\dfrac{2}{3} \;\Rightarrow\; \cos^2\dfrac{x}{2}=\dfrac{1}{3}

\tan\dfrac{x}{2}=\dfrac{\sqrt6/3}{-\sqrt3/3}=-\dfrac{\sqrt6}{\sqrt3}=-\sqrt2

Since x lies in quadrant II, π/2 < x < π, so π/4 < x/2 < π/2 — x/2 lies in quadrant I, where sin, cos and tan are all positive.

\cos^2x=1-\sin^2x=1-\dfrac{1}{16}=\dfrac{15}{16}

In quadrant II, cos x is negative:

2\sin^2\dfrac{x}{2}=1-\cos x=1+\dfrac{\sqrt{15}}{4}=\dfrac{4+\sqrt{15}}{4} \;\Rightarrow\; \sin^2\dfrac{x}{2}=\dfrac{4+\sqrt{15}}{8}

2\cos^2\dfrac{x}{2}=1+\cos x=1-\dfrac{\sqrt{15}}{4}=\dfrac{4-\sqrt{15}}{4} \;\Rightarrow\; \cos^2\dfrac{x}{2}=\dfrac{4-\sqrt{15}}{8}

\tan\dfrac{x}{2}=\sqrt{\dfrac{4+\sqrt{15}}{4-\sqrt{15}}}

Multiply inside the root, top and bottom, by (4 + √15):

\dfrac{4+\sqrt{15}}{4-\sqrt{15}}\times\dfrac{4+\sqrt{15}}{4+\sqrt{15}}=\dfrac{(4+\sqrt{15})^2}{16-15}=(4+\sqrt{15})^2

So the square root simplifies directly: