Question

Please help me figure out how to find the answer?

cos(x+y) if tanx=5/3 and siny=1/3

[answer: (12sqrt17-5sqrt34)/102

Answer 1

I got \[cosy=(2\sqrt3)/2\] and \[tany=\sqrt3/6\] but idk if those are right.

Answer 2

I don't know how to find cosx and sinx from there.

Answer 3

first off.... I assume you know the sum identity for cos(x+y) ?

Answer 4

Yes, \[\cos(x+y)=cosxcosy \pm sinxsiny\]

Answer 5

right

Answer 6

one sec

Answer 7

\(\bf
tan(x)=\cfrac{5}{3}\implies \cfrac{opposite}{adjacent}\implies \cfrac{b}{a}\implies \cfrac{b=5}{a=3}\\ \quad \\
cos(y)=\cfrac{a}{{\color{blue}{ c}}}\qquad c^2=a^2+b^2\implies {\color{blue}{ c}}=\sqrt{a^2+b^2}\\ \quad \\
sin(y)=\cfrac{b}{{\color{blue}{ c}}}\\ \quad \\
------------------------\\
sin(y)=\cfrac{1}{3}\implies \cfrac{opposite}{hypotenuse}\implies \cfrac{b}{c}\implies \cfrac{b=1}{c=3}\\ \quad \\
cos(x)=\cfrac{{\color{blue}{ a}}}{c}\qquad c^2=a^2+b^2\implies \sqrt{c^2-b^2}={\color{blue}{ a}}\)

Answer 8

hmmm shoot, I got my haemm angles a bit.. anyhow \(\bf tan(x)=\cfrac{5}{3}\implies \cfrac{opposite}{adjacent}\implies \cfrac{b}{a}\implies \cfrac{b=5}{a=3}\\ \quad \\
cos(x)=\cfrac{a}{{\color{blue}{ c}}}\qquad c^2=a^2+b^2\implies {\color{blue}{ c}}=\sqrt{a^2+b^2}\\ \quad \\
sin(x)=\cfrac{b}{{\color{blue}{ c}}}\\ \quad \\
------------------------\\
sin(y)=\cfrac{1}{3}\implies \cfrac{opposite}{hypotenuse}\implies \cfrac{b}{c}\implies \cfrac{b=1}{c=3}\\ \quad \\
cos(y)=\cfrac{{\color{blue}{ a}}}{c}\qquad c^2=a^2+b^2\implies \sqrt{c^2-b^2}={\color{blue}{ a}}\)

Answer 9

once you get the missing, using the pythagorean theorem, you'd know the sine and cosine for angles "x" and "y"

once you have those, just plug them in in the cos(x+y) sum identity

Answer 10

I got the answer!! Thank you.

Answer 11

yw

Answer 12

Would you mind doing one more?

sec(x-y) if cscx=5/3 and tany=12/5

Answer 13

tis the same... really, just using the pythagorean theorem to find the missing side(s)

\(\bf csc(x)=\cfrac{5}{3}\implies \cfrac{hypotenuse}{opposite}\implies \cfrac{c}{b}\implies \cfrac{c=5}{b=3}\\ \quad \\
tan(y)=\cfrac{12}{5}\implies \cfrac{opposite}{adjacent}\implies \cfrac{b}{a}\implies \cfrac{b=12}{a=5}\)

Answer 14

just keep in mind that \(\bf sec(\theta)=\cfrac{1}{cos(\theta)}\qquad \qquad sec(x-y)=\cfrac{1}{cos(x-y)}\)