Question

Rita is proving that the following trigonometric identity is true: cos(-��)sin(-��)=-1tan��. Which step would be the first line of her proof?

1) cos��sin��=-1tan��
2) cos(-��)sin��=-1tan��
3) cos(-��) ��� tan��=sin(-��)
4) cos��-sin��=-1tan��

Answer 1

The first line of Rita's proof is 'cos(-α) - tanα = sin(-α)'

Step-by-step explanation:

The first line of Rita's proof would be:

3) cos(-α) - tanα = sin(-α)

Rita starts by applying the trigonometric identity cos(-α) = cosα and sin(-α) = -sinα. Then, she uses the identity cos²α + sin²α = 1 to obtain the equation cos²α - tan²α = 1. Finally, she rearranges the terms to get cos²α - 1 = tan²α and simplifies it to cos(-α) - tanα = sin(-α).

Answer 2

The first line of Rita's proof should demonstrate the application of the even-odd properties of trigonometric functions: cos(-θ) becomes cos(θ) and sin(-θ) becomes -sin(θ). From there, she can simplify and manipulate the expression using the definition of the tangent function to prove the identity.

Step-by-step explanation:

The question at hand involves proving a trigonometric identity that deals with the cosine and sine functions of negative angles and their relation to the tangent function. To begin the proof of the identity cos(-θ)sin(-θ) = -1/tanθ, it is important to apply the even-odd properties of trigonometric functions. Specifically, we know that cosine is an even function, meaning cos(-θ) = cos(θ), and sine is an odd function, which gives us sin(-θ) = -sin(θ). Hence, we can rewrite the initial expression by using these properties.

Therefore, the first line of Rita's proof should show the application of these properties to the identity:

• cos(-θ)sin(-θ) = cos(θ)(-sin(θ))

This can be simplified to -cos(θ)sin(θ). From there, the proof would involve manipulating the expression further by using the definition of the tangent function, which is tan(θ) = sin(θ)/cos(θ), to show that the original identity holds true. As she progresses, she will aim to express the product of cosine and sine in terms of tangent.