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How do you verify # (sin(theta)+cos(theta))^(2) + (sin(theta)-cos(theta))^(2)=2#?

Answer 1

Victoria Adebayo

See below

Left Side #=sin^2theta+2sin theta cos theta + cos^2 theta+sin^2theta-2sin theta cos theta + cos^2 theta#

#=(sin^2 theta +cos^2theta)+(sin^2theta+cos^2theta)#

#=1+1#

#=2#

=Right Side

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Answer 2

Caroline Adams

To verify the identity (sin(theta) + cos(theta))^2 + (sin(theta) - cos(theta))^2 = 2, we can expand each term and then simplify the expression:

(sin(theta) + cos(theta))^2 = sin^2(theta) + 2sin(theta)cos(theta) + cos^2(theta) (sin(theta) - cos(theta))^2 = sin^2(theta) - 2sin(theta)cos(theta) + cos^2(theta)

Adding these two expressions together:

(sin^2(theta) + 2sin(theta)cos(theta) + cos^2(theta)) + (sin^2(theta) - 2sin(theta)cos(theta) + cos^2(theta))

This simplifies to:

sin^2(theta) + sin^2(theta) + cos^2(theta) + cos^2(theta) = 2(sin^2(theta) + cos^2(theta))

Using the trigonometric identity sin^2(theta) + cos^2(theta) = 1:

2(1) = 2

Therefore, (sin(theta) + cos(theta))^2 + (sin(theta) - cos(theta))^2 = 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.