Question

what is the value of 1+sin2x and 1-sin2x

Answer 1

Answer:

$i) 1+sin 2x = (sin x + cos x)^{2}$

$ii) 1-sin 2x = (sin x - cos x)^{2}$

Step-by-step explanation:

$i) 1+sin 2x\\ = 1 + 2sinxcosx$

\* sin 2x = 2sinxcosx *\

$= sin^{2} x + cos^{2} x + 2sinxcosx$

\* By trigonometric identity :

1 = sin² x + cos² x *\

$= (sin x + cos x)^{2}$

\* By algebraic identity:

a² + b² + 2ab = (a+b)² */

$ii) 1-sin 2x \\= sin^{2} x + cos^{2} x - 2sinxcosx \\= (sin x - cos x)^{2}$

Therefore.,

$i) 1+sin 2x = (sin x + cos x)^{2}$

$ii) 1-sin 2x = (sin x - cos x)^{2}$

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Answer 2

1 +sin 2x =(sinx+cosx)^2

1-sin 2x= (sinx-cosx)^2