Question

(sin x+ cos x)^2-2sinx. cosx is equal to
A. 0
B. 2
C. 1
D. sin^2 x-cos^2 x

Answer 1

Hola User________________

Here is Your Answer..!!
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↪Actually welcome to the concept of the TRIGONOMETRY...

↪Basically here applying the formula of (a + b)^2

↪since that is a^2 + b^2 + 2ab

↪So solving here we get as ...

↪Sin^2 x + Cos^2 + 2Sinx .Cosx ( - 2 Sinx . Cosx )

↪solving further the last terms gets cancelled so we remain with ..

↪Sin^2 x + Cos^2 X =》 1
↪since it is identity

↪thus its value is 1
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↪⭐HOPE it helps u..☺

Answer 2

Answer:

C. 1 is the answer.

Step-by-step explanation:

We have to find out (sinx+cosx)2 - 2sin(x)cos(x) = ?

Let us first find out the value of: (sinx+cosx)2

1) Change (sinx+cosx)2 to (sinx+cosx)(sinx+cosx) (since the square of any expression is that expression multiplied by itself.)

2) Utilize the FOIL method for multiplying binomials, e.g. (sinx+cosx)(sinx+cosx)=(sinx)(sinx)+(sinx)(cosx)+(cosx)(sinx)+(cosx)(cosx)

3) Simplify and group like terms: (sinx)(sinx)+(sinx)(cosx)+(cosx)(sinx)+(cosx)(cosx)=sin2x+cos2x+2sinxcosx

4) Recall the trigonometric identity which states sin2x+cos2x=1, and substitute into (3): sin2x+cos2x+2sinxcosx=1+2sinxcosx

5) Use substitution: (sinx+cosx)2=1+2sinxcosx

Now, putting the value of (sinx+cosx)2 in the question;

(sinx+cosx)2 - 2sin(x)cos(x) = 1+2sinxcosx - 2sin(x)cos(x) = 1