i don't even get a tint of this

Question

i don't even get a tint of this <.<

surjithayer · Answer

what the first step can be?

surjithayer · Answer

so i give you hint.

snowflake0531 · Answer

If I put in sina = -4/5 it gives me a decimal as a

surjithayer · Answer

@surjithayer wrote:

$$\sin ^2\alpha+\cos ^2\alpha=1,\cos ^2\alpha=1-\sin ^2 \alpha$$ $$\sec ^2\beta-\tan ^2\beta=1,\tan ^2 \beta=\sec ^2\beta-1$$

surjithayer · Answer

find cos alpha and sec beta

surjithayer · Answer

$$\sec ^2 \beta=1+\tan ^2\beta$$

snowflake0531 · Answer

uh, color me confused
How can I use the squared stuff for this

surjithayer · Answer

take the squareroot to find cos alpha and sec beta

surjithayer · Answer

$$\sin \alpha=-\frac{ 4 }{ 5},\sin ^2\alpha=\frac{ 16 }{ 25 }$$

surjithayer · Answer

$$\beta ~is~in~2nd~quadrant$$

surjithayer · Answer

$$ \alpha ~is~in~4th~quadrant$$

surjithayer · Answer

in 2nd quadrant sin and cosec are positive.
in 4th quadrant sec and cos are positive.

surjithayer · Answer

do you know the formula of cos (x+y) ?

surjithayer · Answer

$$\cos (x+y)=\cos x \cos y-\sin x \sin y$$

snowflake0531 · Answer

sorry
so i got
sina = -4/5, cosa = 3/5

snowflake0531 · Answer

@surjithayer wrote:

do you know the formula of cos (x+y) ?

yes, but I'm still confused about how to get sin and cos b

snowflake0531 · Answer

@surjithayer wrote:

$$\sec ^2 \beta=1+\tan ^2\beta$$

sec^2 b = 225/8 + 1

snowflake0531 · Answer

But I'm confused at what that brings me to

surjithayer · Answer

$$\sec ^2\beta=1+\tan ^2 \beta=1+(\frac{ 225 }{ 64})=\frac{ 289 }{ 64}$$
$$\sec \beta=-\sqrt{\frac{ 289 }{ 64 }}=-\frac{ 17 }{ 8 }$$
$$\cos \beta=\frac{ 1 }{ \sec \beta }=-\frac{ 8 }{ 17 }$$
$$\tan \beta=\frac{ \sin \beta }{ \cos \beta}$$
can you find sin beta?

snowflake0531 · Answer

$$\frac{ -15 }{ 8 } = \frac{ \sin \beta }{ \cos \frac{ -8 }{ 17 } }$$
i don't really understand it that way?
can i just use sin^2 x + cos^2x =1
$$(\frac{ -8 }{ 17 })^2 + \sin^2x =1$$ $$\sin^2x = 1-\frac{ 64 }{ 289 }$$ $$\sin^2x = \frac{ 225 }{ 289 }$$ $$\sin \beta = \frac{ 15 }{ 17 }$$ is that right?

surjithayer · Answer

you are correct.
$$-\frac{ 15 }{ 8 }=\frac{ \sin \beta }{-\frac{ 8 }{ 17} },\sin \beta=-\frac{ 15 }{ 8 }\times \frac{ -8 }{ 17 }=\frac{ 15 }{17 }$$

surjithayer · Answer

now substitute all the values in the formula

snowflake0531 · Answer

okay, altho give me a minute, i need to finish writing all of that on paper e.e

snowflake0531 · Answer

$$\cos(A+B) = sinAcosB+cosAsinB$$
$$(\frac{ -4 }{ 5 })(\frac{ -8 }{ 17 }) + (\frac{ 3 }{ 5 })(\frac{ 15 }{ 17 }) = \frac{ 32 }{ 85 }+\frac{ 45 }{ 85} = \frac{ 77 }{ 85 }$$

snowflake0531 · Answer

is that right? 77/85?

surjithayer · Answer

$$\cos (A+B)=\cos A \cos B-\sin A \sin B$$

snowflake0531 · Answer

oops

snowflake0531 · Answer

then it's
-12/85

snowflake0531 · Answer

wait no
13/85

snowflake0531 · Answer

wait
*-13/85

surjithayer · Answer

$$\cos (\alpha+\beta)=\frac{ 3 }{ 5 }\times \frac{ -8 }{ 17 }-\frac{ -4 }{ 5}\times \frac{ 15 }{ 17 }$$
$$=\frac{ -24+60 }{ 85 }$$
=?

snowflake0531 · Answer

OH i did it totally wrong, i see now e.e

snowflake0531 · Answer

it would be 36/85

surjithayer · Answer

correct

snowflake0531 · Answer

thank you sooooooooooooo muchhhhh

surjithayer · Answer

yw