${sin}^{2} A + {sin}^{2} (A - B) + 2 sin A cos B sin (B - A) =$ $sin^2A+sin^2(A-B)+2sinAcosBsin(B-A)=$ ?

Question

${sin}^{2} A + {sin}^{2} (A - B) + 2 sin A cos B sin (B - A) =$ $sin^2A+sin^2(A-B)+2sinAcosBsin(B-A)=$ ?

options are

${sin}^{2} A$ $sin^2A$

${sin}^{2} B$ $sin^2B$

${cos}^{2} A$ $cos^2A$

${cos}^{2} B$ $cos^2B$

1 Answer

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Answer 1 · 2018-08-10T19:36:42

2.

Explanation:

${sin}^{2} A + {sin}^{2} (A - B) + 2 sin A cos B sin (B - A) =$ $sin^2A+sin^2(A-B)+2sinAcosBsin(B-A)=$

${sin}^{2} a + (sin a cos b - cos a sin b) (sin a cos b - cos a sin b) + 2 sin a cos b (sin b cos a - cos b sin a) =$ $sin^2a+(sinacosb-cosasinb)(sinacosb-cosasinb)+2sinacosb(sinbcosa-cosbsina)=$

${sin}^{2} a + {sin}^{2} a {cos}^{2} b - 2 sin a cos a sin b cos b + {cos}^{2} a {sin}^{2} b + 2 sin a sin b cos a cos b - 2 {cos}^{2} b {sin}^{2} a =$ $sin^2a+sin^2acos^2b-2sinacosasinbcosb+cos^2asin^2b+2sinasinbcosacosb-2cos^2bsin^2a=$

${sin}^{2} a - {sin}^{2} a {cos}^{2} b + {cos}^{2} a {sin}^{2} b =$ $sin^2a-sin^2acos^2b+cos^2asin^2b=$

${sin}^{2} a (1 - {cos}^{2} b + {cot}^{2} a {sin}^{2} b) =$ $sin^2a(1-cos^2b+cot^2asin^2b)=$

${sin}^{2} a {sin}^{2} b (1 + {cot}^{2} a) =$ $sin^2asin^2b(1+cot^2a)=$

${sin}^{2} a {sin}^{2} b ({csc}^{2} a) =$ $sin^2asin^2b(csc^2a)=$

${sin}^{2} a {sin}^{2} b (\frac{1}{{sin}^{2} a}) =$ $sin^2asin^2b(1/sin^2a)=$

${sin}^{2} b$ $sin^2b$

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${sin}^{2} A + {sin}^{2} (A - B) + 2 sin A cos B sin (B - A) =$ $sin^2A+sin^2(A-B)+2sinAcosBsin(B-A)=$ ?

options are

${sin}^{2} A$ $sin^2A$

${sin}^{2} B$ $sin^2B$

${cos}^{2} A$ $cos^2A$

${cos}^{2} B$ $cos^2B$

1 Answer

Explanation:

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Math

Humanities

... and beyond

sin2A+sin2(A−B)+2sinAcosBsin(B−A)=sin^2A+sin^2(A-B)+2sinAcosBsin(B-A)= ?

options are sin2Asin^2A sin2Bsin^2B cos2Acos^2A cos2Bcos^2B

1 Answer

Explanation:

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${sin}^{2} A + {sin}^{2} (A - B) + 2 sin A cos B sin (B - A) =$ $sin^2A+sin^2(A-B)+2sinAcosBsin(B-A)=$ ?

options are

${sin}^{2} A$ $sin^2A$

${sin}^{2} B$ $sin^2B$

${cos}^{2} A$ $cos^2A$

${cos}^{2} B$ $cos^2B$