Question

Trigonometry Help

Find the exact value of cos(x + y) if sinx = 4/5 and siny = 12/13, with x in quadrant II and y in quadrant II.

cos(x + y) = ?
**Simplify your answer. Type a fraction**

Answer 1

so and "x" and "y" angle are both in the 2nd quadrant

what sign are cosine and sine in the 2nd quadrant?

both positive? both negative? or one negative and the other positive?

Answer 2

(negative, positive)

Answer 3

sorry i started working another problem while i waited for someone to come help me.

Answer 4

so on quadrant II sine is positive and cosine is negative

notice the given ones, are both sine, both are positive
so what's their cosine?
well

\(\bf sin(x)=\cfrac{4}{5}\to \cfrac{opposite}{hypotenuse}\to \cfrac{b=4}{c=5}
\\ \quad \\
c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}={\color{blue}{ a}}\qquad cos(x)=-\cfrac{{\color{blue}{ a}}}{c}\)

since the cosine is in the II quadrant, we'll use the negative version from the pythagorean theorem
------------------------------------------

\(\bf sin(y)=\cfrac{12}{13}\to \cfrac{opposite}{hypotenuse}\to \cfrac{b=12}{c=13}
\\ \quad \\
c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}={\color{blue}{ a}}\qquad cos(y)=-\cfrac{{\color{blue}{ a}}}{c}\)

Answer 5

and of course, recall that \(\bf cos(x+y)=cos(x)cos(y)-sin(x)sin(y)\)

Answer 6

I think so... BUT recall -> **Simplify your answer. Type a fraction**

you're meant to keep it as fraction

Answer 7

ooh you're right, i forgot about that. thanks. ill take another shot at it

Answer 8

\(\bf sin(x)=\cfrac{4}{5}\to \cfrac{opposite}{hypotenuse}\to \cfrac{b=4}{c=5}
\\ \quad \\
c^2=a^2+b^2\implies \pm \sqrt{5^2-4^2}={\color{blue}{ a}}\implies -\sqrt{9}={\color{blue}{ a}}
\\ \quad \\
-3={\color{blue}{ a}}\qquad cos(x)=-\cfrac{{\color{blue}{ -3}}}{5}\)

-------------------------------------------------------
\(\bf sin(y)=\cfrac{12}{13}\to \cfrac{opposite}{hypotenuse}\to \cfrac{b=12}{c=13}
\\ \quad \\
c^2=a^2+b^2\implies \pm \sqrt{13^2-12^2}={\color{blue}{ a}}\implies -\sqrt{25}={\color{blue}{ a}}
\\ \quad \\
-5={\color{blue}{ a}}\qquad cos(y)=-\cfrac{{\color{blue}{ -5}}}{13}\)

Answer 9

wwait why are you getting a negative for a?

Answer 10

well... the fractions look ok, though you're missing the negative part, yes

recall that on the II Quadrant, cosine is negative

Answer 11

also keep in mind that, the fractions ARE the cosine value, so you don't need to wrap them in cos() or sin()

Answer 12

ooooh i see, that makes sense now,

Answer 13

so its that answer without the cos and sin

Answer 14

yes, just the fractions

Answer 15

so where do we go from there? or that is the end of the probelm?

Answer 16

\(\bf cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
\\ \quad \\
cos(x+y)=\cfrac{{\color{blue}{ -3}}}{5}\cdot \cfrac{{\color{blue}{ -5}}}{13}-\cfrac{4}{5}\cdot \cfrac{12}{13}\)

Answer 17

alright i got that, should i put it all over a common denominator?

Answer 18

well, is really just a subtraction.... so you combine the factors and then subtract, leave it as fraction :)

Answer 19

need to dash, but I'll be here tomorrow :)

Answer 20

oh no problem thanks for all youre hlp!