0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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, $\cos (a+b) = (\cos a \cos b) - (\sin a \sin b).$ $\endgroup$ Nov 9 '20 at 5:17
0
$\begingroup$

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$ $\sin(a)=\dfrac 3 5$ 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.

$\endgroup$
0
$\begingroup$

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 $a-b$)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.