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Find the value of the expression 3{sin^...

Find the value of the expression `3{sin^4((3pi)/2-theta)+sin^4(3pi+theta)}-2{sin^6(pi/2+theta)+sin^6(5pi-theta)}`

Text Solution

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Given,
`3[sin ^{4}(frac{3 pi}{2}-alpha)+sin ^{4}(3 pi+alpha)]-2[sin ^{6}(frac{pi}{2}+alpha)-sin ^{6}(5 pi-alpha)]`

`=3[cos ^{4} alpha+sin ^{4} alpha]-2[cos ^{6} alpha+sin ^{6} alpha]`
`=3[(cos ^{2} alpha)^{2}+(sin ^{2} alpha)^{2}]-2[(sin ^{2} alpha)^{3}+(cos ^{2} alpha)^{3}]`
`=3[(sin ^{2} alpha+cos ^{2} alpha)^{2}-2 sin ^{2} alpha cos ^{2} alpha]-2[(sin ^{2} alpha+cos ^{2} alpha)(sin ^{4} alpha+cos ^{4} alpha-sin ^{2} alpha cos ^{2} alpha)]`
Using `a^{4}+b^{4}=(a^{2}+b^{2})-2 a^{2} b^{2}`
and `a^6+b^{6}=(a^{2})^{3}+(b^{2})^{3}=(a^{2}+b^{2})(a^{4}+b^{4}-a^{2} b^{2})`
`=3[1-2 sin ^{2} alpha cos ^{2} alpha]-2[(1)((sin ^{2} alpha+cos ^{2} alpha)^{2}-2 sin ^{2} alpha cos ^{2} alpha-sin ^{2} alpha cos ^{2} alpha)]`
...
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Knowledge Check

  • cos^2 (pi/6 + theta) - sin^2 (pi/6 - theta) =

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