$\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$

Question

It is known that $$\tan(a) = \frac{3}{4}, \:\:\: \tan(b) = \frac{5}{12} $$ with $a,b < \frac{\pi}{2}$. What is $\cos(a+b)$?

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Attempt :

$$ \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$ And we can write $\tan(a) = \sin(a)/\cos(a) = 0.3/0.4 $ and $ \sin(b)/\cos(b) = 0.05/0.12 $ so $$ \cos(a+b) = (0.4)(0.12) - (0.3) (0.05) = 33/1000$$

Is this correct? Thanks.

Answer 1

No, this can't be correct. Remember that $\sin^2x+\cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.

It turns out that $$\sin a=\frac35\qquad\cos a=\frac45$$ $$\sin b=\frac5{13}\qquad\cos b=\frac{12}{13}$$ and thus $$\cos(a+b)=\cos a\cos b-\sin a\sin b=\frac45\cdot\frac{12}{13}-\frac35\cdot\frac5{13}=\frac{33}{65}$$

Answer 2

Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):

$sin(a)=\frac{3}{5}$

$cos(a)=\frac{4}{5}$

$sin(b)=\frac{5}{13}$

$cos(b)=\frac{12}{13}$

Thus $$cos(a+b)=\frac{4}{5}\times\frac{12}{13}-\frac{3}{5}\times\frac{5}{13}=\frac{33}{65}$$

Answer 3

$$ \tan(a+b)= \frac {\tan a +\tan b}{1-\tan a \tan b} = \frac {3/4 +5/12}{1-(3/4)(5/12)}= \frac {56}{33}$$

$$ \sec^2(a+b)=1+\tan^2(a+b)=\frac {4225}{1089}=(\frac {65}{33})^2$$

$$\cos(a+b)= 33/65$$