# Calculate the values of the "trigonometric" functions for the angles $a+b$ and $a-b$

Calculate the values of the "trigonometric" functions for the angles $$a+b$$ and $$a-b$$ if $$\sin a =\frac{3}{5} \, y\, \sin b= \frac{2\sqrt{13}}{13}$$

I did for $$\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)= \frac{3}{5}\cos(b)+\frac{2\sqrt{13}}{13}\cos(a)$$

but I don't know if am I the correct way and how to know how much is $$\cos(b)$$ and $$\cos(a)$$ how can I calculate it? just with calculator?

• Use the formula $\sin^2 \theta + \cos^2 \theta = 1.$ Presumably, you are allowed to assume that $\cos a$ and $\cos b$ are both positive. Without an assumption like this, the problem can't be solved. Also, $\cos (a+b) = (\cos a \cos b) - (\sin a \sin b).$ Nov 9 '20 at 5:17

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$ $$\sin(a)=\dfrac 3 5$$ implies a right triangle with hypotenuse $$5$$ and side opposite $$a$$ of $$3$$

$$\therefore$$ side adjacent to $$a$$ is $$4$$ $$\implies \cos(a)=\dfrac 4 5$$

Now, $$\dfrac {2\sqrt{13}} {13}=\dfrac 2 {\sqrt{13}}$$, implying a hypotenuse of $$\sqrt{13}$$, opposite of $$2$$, and by Pythagoras, adjacent of $$3$$.

$$\therefore \cos(b)=\dfrac 3 {\sqrt{13}}$$

Substitution into angle formula and simplifying gives $$\sin(a+b)=\dfrac {17\sqrt{13}} {65}$$

I'll leave $$\sin(a-b)$$ to you.

There are 3 concepts (which I am assuming you already know):

1.) Definition of $$\tan x$$ and $$\cot x$$ in terms of $$\sin x$$ and $$\cos x$$

2.) Reciprocal-relations like $$\sin{x}=\frac{1}{\operatorname{cosec}x}$$

3.) Trigonometric identities which are based on Pythagoras theorem like $$\sin^2 {x}+\cos^2{x}=1$$ (note this identity was proved using Pythagoras theorem)

Now, we have 3 such reciprocal relations and 3 such identities based on the Pythagoras theorem (there are more trigonometric identities based on the Pythagoras theorem, but we can generally derive all of them from these 3 fundamental identities).

The best part of those above concepts is that they let us find any trigonometric ratio easily if we already know one. As:

If $$\sin{x}=\frac{1}{2}$$ then, $$\operatorname{cosec} x=\frac{1}{\sin x}=2$$

and, $$\sin^2 {x}+\cos^2{x}=1$$ $$\Rightarrow\cos^2{x}=1-\frac{1}{4}$$ $$\Rightarrow \cos^2 x= \frac{3}{4} \Rightarrow \cos x=\pm \frac {\sqrt3}{2}$$

Now, if we have this angle $$x$$ only defined for a right-angled triangle, then we can say that $$\cos x \neq -\frac {\sqrt 3}{2}$$, since trigonometric ratios of angle $$x$$ where $$0°\lt x \lt 90°$$, are always positive. So, that gives us $$\cos x= \frac {\sqrt3}{2}$$

However, if $$x$$ is defined for any angle then both values of $$\cos x$$ is right. So, $$\cos x= \pm \frac{\sqrt 3}{2}$$

Now we can find $$\sec x$$ by reciprocal relation and $$\tan x$$ and $$\cot x$$ by their definition.

If we have taken $$x$$ as any angle, you can observe $$x$$ belongs from 1st or 2nd quadrant, since $$\sin x$$ which is $$\frac {1}{2}$$, is positive. So, we will get $$\operatorname {cosec} x$$ positive and $$\cos x$$, $$\sec x$$, $$\tan x$$ and $$\cot x$$ either positive or negative.

I also want to suggest that once you have found one trigonometric ratio of $$a+b$$, use that ratio to find other ratios of $$a+b$$ rather than writing compound angle formulae for each trigonometric ratio of $$a+b$$. (Similar with $$a-b$$)