Evaluate : sin^6 theta - cos^6 theta
Answers
We can first try factoring a3−b3=(a−b)(a2+ab+b2) . With a=sinθ and b=cos2θ , the first factor becomes −cos2θ and the second factor is
sin4θ+sin2θcos2θ+cos4θ
Now x4+y4=(x2+y2)2−2x2y2 , so the factor above becomes
(sin2θ+cos2θ)2−sin2θcos2θ=1−14sin22θ
so a possible answer is
14sin22θcos2θ−cos2θ=18sin4θsin2θ−cos2θ
that can be further reduced to
116cos2θ−116cos6θ−cos2θ
and finally to
−1516cos2θ−116cos6θ
How do I evaluate sin^6 theta-cos^6 theta?
Use class 11th trigonometry formula. Firstly make sin and cos in square forms and square power raised to 3.
Rest is your work this hint is enough for you. Such questions are there in RD sharma and KC sharma. Practice more similar questions
If sinθ+sin2θ=1, then evaluate cos2θ+cos4θ?
5cos2θ+7sin2θ=6 where theta is acute then can you evaluate tanθ ?
5cos2θ+7sin2θ=6 (Given)
tanθ>0 (Given: θ is acute)
cos2θ+sin2θ=1 (Trigonometric identity)
5cos2θ+5sin2θ=5 (Multiply both sides of #3 by 5 )
2sin2θ=1 (Subtract #4 from #1)
sin2θ=12 (Divide sides of #5 by 2 )
cos2θ+12=1 (Substitute #6 into left side of #3)
cos2θ=12 (Subtract 12 from sides of #7)
tanθ=sinθcosθ (Trigonometric identity)
tan2θ=sin2θcos2θ (Square sides of #9)
tan2θ=1/21/2=1 (Substitute #6 and #8 into right side of #10)
tanθ=1 (Take square roots of #11, noting #2)
How do I evaluate sin^6 theta-cos^6 theta?
sin^6 x - cos^6 x = (sin²x)³ - (cos²x)³.
Now using a³-b³= (a-b)(a²+b²+ab) we get,
LHS = (sin²x - cos²x)(sin²x + cos²x + sin x. cos x)
= {-(cos²x-sin²x)}(1+ sin x cos x)
If you have not studied the double angle formulas plz don't read further. This is the answer in that case.
= (-cos 2x)(1+ sin2x/2) Ans.
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