$2 (sin^{6} θ + cos^{6} θ) - 3 (sin^{4} θ + cos^{4} θ) + 1$ is equal to

2

0

4

6

Views: 855 students

Found 8 tutors discussing this question

Theodore Discussed

2 (sin^{6} θ + cos^{6} θ) - 3 (sin^{4} θ + cos^{4} θ) + 1

is equal to

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

2 (sin^{6} θ + cos^{6} θ) - 3 (sin^{4} θ + cos^{4} θ) + 1

Using (a^3+b^3=(a+b)^3-3\cdot a\cdot b(a+b)) \ and \ (a^2+b^2=(a+b)^2-2ab) , we get

= 2 ((sin^{2} θ + cos^{2} θ)^{3} - 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ)) - 3 ((sin^{2} θ + cos^{2} θ)^{2} - 2 sin^{2} θ cos^{2} θ) + 1

= 2 - 6 sin^{2} θ cos^{2} θ - 3 + 6 sin^{2} θ cos^{2} θ + 1 = 0

Was this solution helpful?

150

Report

One destination to cover all your homework and assignment needs

Learn Practice Revision Succeed

Instant 1:1 help, 24x7

60, 000+ Expert tutors

Textbook solutions

Big idea maths, McGraw-Hill Education etc

Essay review

Get expert feedback on your essay

Schedule classes

High dosage tutoring from Dedicated 3 experts

Trusted by 4 million+ students

Stuck on the question or explanation?

Connect with our math tutors online and get step by step solution of this question.

231 students are taking LIVE classes

Question Text	$2 (sin^{6} θ + cos^{6} θ) - 3 (sin^{4} θ + cos^{4} θ) + 1$ is equal to
Answer Type	Text solution:1
Upvotes	150