Math  /  Trigonometry

QuestionProve that cos2ϕ1sinϕ=1+sinϕ\frac{\cos ^{2} \phi}{1-\sin \phi}=1+\sin \phi, where sinϕ1\sin \phi \neq 1, by expressing cos2ϕ\cos ^{2} \phi in terms of sinϕ\sin \phi.

Studdy Solution

STEP 1

1. We are given the trigonometric identity to prove: cos2ϕ1sinϕ=1+sinϕ\frac{\cos^{2} \phi}{1-\sin \phi} = 1+\sin \phi.
2. We will use the Pythagorean identity to express cos2ϕ\cos^{2} \phi in terms of sinϕ\sin \phi.
3. We assume sinϕ1\sin \phi \neq 1 to avoid division by zero.

STEP 2

1. Express cos2ϕ\cos^{2} \phi in terms of sinϕ\sin \phi.
2. Substitute the expression for cos2ϕ\cos^{2} \phi into the left-hand side of the given identity.
3. Simplify the expression to show it equals the right-hand side.

STEP 3

Use the Pythagorean identity to express cos2ϕ\cos^{2} \phi in terms of sinϕ\sin \phi:
cos2ϕ=1sin2ϕ \cos^{2} \phi = 1 - \sin^{2} \phi

STEP 4

Substitute cos2ϕ=1sin2ϕ\cos^{2} \phi = 1 - \sin^{2} \phi into the left-hand side of the identity:
cos2ϕ1sinϕ=1sin2ϕ1sinϕ \frac{\cos^{2} \phi}{1-\sin \phi} = \frac{1 - \sin^{2} \phi}{1-\sin \phi}

STEP 5

Notice that the numerator 1sin2ϕ1 - \sin^{2} \phi can be factored as a difference of squares:
1sin2ϕ=(1sinϕ)(1+sinϕ) 1 - \sin^{2} \phi = (1 - \sin \phi)(1 + \sin \phi)
Substitute this back into the expression:
1sin2ϕ1sinϕ=(1sinϕ)(1+sinϕ)1sinϕ \frac{1 - \sin^{2} \phi}{1-\sin \phi} = \frac{(1 - \sin \phi)(1 + \sin \phi)}{1-\sin \phi}

STEP 6

Cancel the common factor of 1sinϕ1 - \sin \phi in the numerator and the denominator:
(1sinϕ)(1+sinϕ)1sinϕ=1+sinϕ \frac{(1 - \sin \phi)(1 + \sin \phi)}{1-\sin \phi} = 1 + \sin \phi

STEP 7

The simplified expression 1+sinϕ1 + \sin \phi matches the right-hand side of the given identity, thus proving the identity:
cos2ϕ1sinϕ=1+sinϕ \frac{\cos^{2} \phi}{1-\sin \phi} = 1 + \sin \phi
The identity is proven.

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