Trig Functions: Given sin x = 6/7, cos x > 0, sin y = -2/

[snakeyesxlaw]

Suppose sin x = 6/7, cos x > 0, sin y = - 2/5, and cos y < 0. Find each of the following quantities:

sin (x +y) =

cos (x + y) =

tan(x + y) =

what i came up with,

sin (x + y) = (6/7) cos y + (-2/5) cos x

cos (x + y) = cos (x) cos (y) + (12/35)

tan (x + y) = any suggestions? please work this particular problem thru

something is wrong?

 

[stapel]

For x, you are given that sin(x) = 6/7. Use the Pythagorean Theorem to find the length of the third side of this first triangle. Also, you are given that sin(x) and cos(x) are positive, so you should know in which quadrant is x, and what are the signs on sine, cosine, and tangent.

You can do the same thing with y. :idea:

What is your question on the tangent angle sum? Didn't they give you an identity for that, too? And even if not, isn't tangent just the sine divided by the cosine? :wink:

Eliz.

 

[skeeter]

you need this identity ... \(\displaystyle \L \sin^2{t} + \cos^2{t} = 1\), and you have to pay attention to the quadrant where angles x and y reside.

also, note that \(\displaystyle \L \tan(x+y) = \frac{\sin(x+y)}{\cos(x+y)}\)

 

[pka]

From the given we can conclude that:
\(\displaystyle \L
\cos (x) = \frac{{\sqrt {13} }}{7}\,\& \,\cos (y) = \frac{{ - \sqrt {21} }}{5}\)