All Exams Test series for 1 year @ ₹349 only
Question

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

This question was previously asked in
HTET 2025 Level 1 PRT Question Paper (5-Jul-2026)
The correct answer is

1/(2√2)

Since \(3\pi/8 = \pi/2 - \pi/8\), we have \(\cos(3\pi/8) = \sin(\pi/8)\) and \(\sin(3\pi/8) = \cos(\pi/8)\).

Substituting, the expression becomes \(\cos^3(\pi/8)\sin(\pi/8) + \sin^3(\pi/8)\cos(\pi/8)\).

Factoring out \(\sin(\pi/8)\cos(\pi/8)\): \(= \sin(\pi/8)\cos(\pi/8)\left[\cos^2(\pi/8)+\sin^2(\pi/8)\right] = \sin(\pi/8)\cos(\pi/8)\), since \(\cos^2\theta+\sin^2\theta=1\).

Using the double angle identity, \(\sin(\pi/8)\cos(\pi/8) = \tfrac{1}{2}\sin(\pi/4) = \tfrac{1}{2}\cdot\tfrac{\sqrt2}{2} = \dfrac{1}{2\sqrt2}\).

Was this answer helpful?

Similar Questions

  1. Value of \(\cot\left(\dfrac{\pi}{20}\right)\cot\left(\dfrac{3\pi}{20}\right)\cot\left(\dfrac{5\pi}{20}\right)\cot\left(\dfrac{7\pi}{20}\right)\cot\left(\dfrac{9\pi}{20}\right)\)

  2. The average of \(2\sin 2^\circ,4\sin 4^\circ,6\sin 6^\circ,\ldots,180\sin 180^\circ\) is:

  3. \(\sin^2 5^\circ+\sin^2 10^\circ+\sin^2 15^\circ+\ldots+\sin^2 90^\circ\) value of:


Important Questions from Trigonometric Identities

  1. What is cos 2β equal to ?

  2. What is the value of sec2γ?

  3. On simplifying \(\frac{{{{\sin }^3}{\rm{A}} + \sin 3{\rm{\;A}}}}{{\sin {\rm{A}}}} + \frac{{{{\cos }^3}{\rm{A}} - \cos 3{\rm{\;A}}}}{{\cos {\rm{A}}}}\) we get

  4. (1 – sin A + cos A) 2is equal to

  5. What is \(\frac{{\cos {\rm{\theta }}}}{{1 - \tan {\rm{\theta }}}} + \frac{{\sin {\rm{\theta }}}}{{1 - \cot {\rm{\theta }}}}\) equal to?

Need Expert Advice?
Upcoming Exams
CTET
September 06, 2026
MH SET
September 06, 2026
Test Series
HTET img
Teaching
HTET TGT Physical Education Mock Test Series
83 Tests 2 Tests Free
889 Attempts
4.2(9)
English, Hindi

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App