Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
301 views
in Trigonometry by (45 points)
The value of \( \left(\sin ^{8} \frac{5 \pi}{12}-\cos ^{8} \frac{5 \pi}{12}\right) \) is

Please log in or register to answer this question.

2 Answers

+1 vote
by (155 points)

Consider the problem

+1 vote
by (53.7k points)

\(sin^8 \left(\frac{5\pi}{12}\right) - cos^8 \left(\frac {5\pi}{12}\right) = ?\)

\(\frac{5\pi}{12} = \frac{5 \times 180°}{12} = 5\times 15° = 75° \)

\(sin\left(\frac{5\pi}{12}\right) = sin75° + sin(30° + 45°)\)

\(= sin30° cos45° + cos30° sin45°\)

\( = \frac12 \times \frac 1{\sqrt2} + \frac{\sqrt 3}2 \times \frac 1{\sqrt 2}\)

\(= \frac 1{2\sqrt 2} ( 1 + \sqrt 3)\)

\(cos\left(\frac {5\pi}{12}\right) = cos 75° = cos(30° + 45°)\)

\(= cos 30° cos 450° - sin30° sin45°\)

\(= \frac {\sqrt 3}2 \times \frac 1{\sqrt 2} - \frac 12 \times \frac1{\sqrt 2}\)

\( = \frac{\sqrt 3 - 1}{2\sqrt 2}\)

\(\therefore sin^8\left(\frac {5\pi}{12}\right) - cos^8 \left(\frac {5\pi}{12}\right) = \left(\frac{\sqrt 3 + 1}{2\sqrt 2}\right)^8 - \left(\frac{\sqrt 3 - 1}{2\sqrt 2}\right)^8\)

\(= \left(\frac 1{2\sqrt 2}\right)^8 \left((\sqrt 3 + 1)^8 - (\sqrt 3 - 1)^8\right)\)

\(= \frac 1{2^8 .2^4} \left(2(8.(\sqrt 3)^7 +56(\sqrt3)^5 +56(\sqrt 3)^3 + 8\sqrt 3)\right)\)

\(= \frac 1{2^{12}} . 16 ((\sqrt 3)^7 + 7(\sqrt 3)^5 + 7(\sqrt 3)^3 + \sqrt 3)\)

\(= \frac {\sqrt 3}{256} ((\sqrt 3)^6 + 7(\sqrt 3)^4 + 7(\sqrt 3)^2 + 1)\)

\(= \frac{\sqrt 3}{256}(27 + 63 + 21 + 1)\)

\(= \frac{\sqrt 3 \times 112}{256}\)

\(= \frac{7\sqrt 3}{16}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...