Question

व्यंजक `3[sin^4 ((3pi)/2 - alpha) + sin^4 (3pi + alpha)] - 2[sin^6 (pi/2 + alpha) + sin^6 (5pi - alpha)]` का मान ज्ञात कीजिए।

Answer 1

`3[sin^4 ((3pi)/2 - alpha) + sin^4 (3pi + alpha)] - 2[sin^6 (pi/2 + alpha) + sin^6 (5pi - alpha)]` का मान ज्ञात कीजिए,

= `3[sin^4((3pi)/2 - alpha) + sin^4(3pi + alpha)] - 2[sin^6(pi/2 - alpha) + sin^6(5pi + alpha)]`

= `3[cos^4alpha + sin^4(pi + alpha)] - 2[cos^6alpha + sin^6(pi + alpha)]`

= `3[cos^4alpha + sin^4alpha + 2sin^2alphacos^2alpha - 2sin^2alphacos^2alpha] - 2[(cos^2alpha + sin^2alpha)^3 - 3sin^2alpha cos^2alpha(cos^2alpha + sin^2alpha)]`

`sin^2theta + cos^2theta = 1` में रखने पर

= `3[(sin^2alpha + cos^2alpha)^2 - 2sin^2alpha cos^2alpha] - 2[1 - 3sin^2alpha cos^2alpha]`

= `3[1 - 2sin^2alpha cos^2alpha] - 2[1 - 3sin^2alpha cos^2alpha]`

= `3 - 6sin^2alpha cos^2alpha - 2 + 6sin^2alpha cos^2alpha`

= 3 - 2

= 1

दी हुई अभिव्यक्ती का मान 1 है।