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Math  /  Trigonometry

Question(sin⁡x+cos⁡x)2=1+2sin⁡xcos⁡x(\sin x+\cos x)^{2}=1+2 \sin x \cos x

Studdy Solution

STEP 1

1. The equation (sin⁡x+cos⁡x)2=1+2sin⁡xcos⁡x(\sin x + \cos x)^2 = 1 + 2 \sin x \cos x is a trigonometric identity.
2. We will use trigonometric identities to simplify and verify the equation.

STEP 2

1. Expand the left side of the equation.
2. Simplify the expanded expression using trigonometric identities.
3. Verify that both sides of the equation are equal.

STEP 3

Expand the left side of the equation (sin⁡x+cos⁡x)2(\sin x + \cos x)^2:
(sin⁡x+cos⁡x)2=(sin⁡x)2+2sin⁡xcos⁡x+(cos⁡x)2(\sin x + \cos x)^2 = (\sin x)^2 + 2 \sin x \cos x + (\cos x)^2

STEP 4

Use the Pythagorean identity (sin⁡x)2+(cos⁡x)2=1(\sin x)^2 + (\cos x)^2 = 1 to simplify the expression:
(sin⁡x)2+2sin⁡xcos⁡x+(cos⁡x)2=1+2sin⁡xcos⁡x(\sin x)^2 + 2 \sin x \cos x + (\cos x)^2 = 1 + 2 \sin x \cos x

STEP 5

Verify that both sides of the equation are equal:
The left side simplifies to:
1+2sin⁡xcos⁡x1 + 2 \sin x \cos x
The right side is already:
1+2sin⁡xcos⁡x1 + 2 \sin x \cos x
Since both sides are equal, the equation is verified as a trigonometric identity.
The equation (sin⁡x+cos⁡x)2=1+2sin⁡xcos⁡x(\sin x + \cos x)^2 = 1 + 2 \sin x \cos x is verified as true.

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