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

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

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.

• You cannot do that: $$\frac{\sin(a)}{\cos(a)}=\frac{0.3}{0.4}\Rightarrow \sin(a)=0.3, \cos(a)=0.4$$. In fact, $\sin^{2}(a)+cos^{2}(a)=0.09+0.16\neq 1$ – Mateus Rocha Jul 17 '18 at 1:03
• This is not since $sin^2(a)+cos^2(a)=1$ – abc... Jul 17 '18 at 1:05
• You can solve for sin(a)(=x) and cos(a)(=y): $\frac{x}{y}=\frac{3}{4}, x^2+y^2=1$ – abc... Jul 17 '18 at 1:05
• You have $\frac {\sin a}{\cos a} = \frac 34$ and $\frac {\sin b}{\cos b} = \frac 5{12}$. Why would you think that means $\cos b\cos b = 4*12*.001$? – fleablood Jul 17 '18 at 1:17
• The 3 answers that have appeared so far have assumed that a,b are positive, but that is not in the Q. Did you mean to assume $a,b$ are positive? If not, the solution for $\cos (a+b)$ is not unique. – DanielWainfleet Jul 17 '18 at 1:53

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 $$\cos(a+b)=\cos a\cos b-\sin a\sin b=\frac45\cdot\frac{12}{13}-\frac35\cdot\frac5{13}=\frac{33}{65}$$

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

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

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

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