The value of \cos ^3( \pi \over 8 )\cos( 3\pi \over 8 ) + \sin ^3( \pi \over 8 )\sin( 3\pi \over 8 ) is :

Trigonometric Ratio and Identites PYQ — JEE Main 2020 · 9 Jan · Shift 1 · Q38 with Solution

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Correct answer: D

Let $a=\frac{\pi}{8}.$ Then $\frac{3\pi}{8}=\frac{\pi}{2}-\frac{\pi}{8}=\frac{\pi}{2}-a.$ So, $\cos\left(\frac{3\pi}{8}\right)=\cos\left(\frac{\pi}{2}-a\right)=\sin a,$ $\sin\left(\frac{3\pi}{8}\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a.$
Substitute into the given expression: $\cos^3\left(\frac{\pi}{8}\right)\cos\left(\frac{3\pi}{8}\right)+\sin^3\left(\frac{\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)$ becomes $\cos^3 a\,\sin a+\sin^3 a\,\cos a.$
Take common factor $\sin a\cos a$ :

=\sin a\cos a(\cos^2 a+\sin^2 a).$$ Using $$\sin^2 a+\cos^2 a=1,$$ we get $$\cos^3 a\sin a+\sin^3 a\cos a=\sin a\cos a.$$ 4. Now use $$\sin a\cos a=\frac{1}{2}\sin 2a.$$ Since $a=\frac{\pi}{8}$, $$\sin a\cos a=\frac{1}{2}\sin\left(\frac{\pi}{4}\right) =\frac{1}{2}\cdot\frac{1}{\sqrt2} =\frac{1}{2\sqrt2}.$$ 5. Therefore, the value is $$\boxed{\frac{1}{2\sqrt2}}.$$ 6. Checking options: - A: $\frac{1}{\sqrt2}$ - B: $\frac{1}{2}$ - C: $\frac{1}{4}$ - D: $\frac{1}{2\sqrt2}$ So the correct option is **D**.

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Correct answer: D

Let $a=\frac{\pi}{8}.$ Then $\frac{3\pi}{8}=\frac{\pi}{2}-\frac{\pi}{8}=\frac{\pi}{2}-a.$ So, $\cos\left(\frac{3\pi}{8}\right)=\cos\left(\frac{\pi}{2}-a\right)=\sin a,$ $\sin\left(\frac{3\pi}{8}\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a.$
Substitute into the given expression: $\cos^3\left(\frac{\pi}{8}\right)\cos\left(\frac{3\pi}{8}\right)+\sin^3\left(\frac{\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)$ becomes $\cos^3 a\,\sin a+\sin^3 a\,\cos a.$
Take common factor $\sin a\cos a$ :

=\sin a\cos a(\cos^2 a+\sin^2 a).$$ Using $$\sin^2 a+\cos^2 a=1,$$ we get $$\cos^3 a\sin a+\sin^3 a\cos a=\sin a\cos a.$$ 4. Now use $$\sin a\cos a=\frac{1}{2}\sin 2a.$$ Since $a=\frac{\pi}{8}$, $$\sin a\cos a=\frac{1}{2}\sin\left(\frac{\pi}{4}\right) =\frac{1}{2}\cdot\frac{1}{\sqrt2} =\frac{1}{2\sqrt2}.$$ 5. Therefore, the value is $$\boxed{\frac{1}{2\sqrt2}}.$$ 6. Checking options: - A: $\frac{1}{\sqrt2}$ - B: $\frac{1}{2}$ - C: $\frac{1}{4}$ - D: $\frac{1}{2\sqrt2}$ So the correct option is **D**.