Given: \(\sin a = -\dfrac{21}{29},\dfrac{3\pi}{2}<a<2\pi\\ \tan b = -\dfrac{24}{7},\dfrac{\pi}{2}<b<\pi\)
Note that: \(\boxed{\color{BurntOrange}{\cos(a + b) = \cos a \cos b - \sin a \sin b}}\)
So that we have to find cos a, cos b, sin b because the value of sin a is given.
A few formulae: \(\cos(\arcsin x) = \sqrt{1-x^2}\\ \cos(\arctan x) = \dfrac{1}{\sqrt{1+x^2}}\\ \sin(\arctan x) = \dfrac{x}{\sqrt{1+x^2}}\)
To find cos a:
\(\cos a\\ = \cos(\arcsin(\sin a))\\ =\sqrt{1-\sin^2 a}\\ =\sqrt{1-\left(-\dfrac{21}{29}\right)^2}\\ =\sqrt{\dfrac{29^2 - 21^2}{29^2}}\\ =\sqrt{\dfrac{20^2}{29^2}}\\ =\dfrac{20}{29}\)
To find cos b:
\(\cos b\\ =\cos(\arctan(\tan b))\\ =\dfrac{1}{\sqrt{1+\tan^2 b}}\\ =\dfrac{1}{\sqrt{1+\dfrac{576}{49}}}\\ =\dfrac{1}{\sqrt{\dfrac{625}{49}}}\\ =\dfrac{1}{\dfrac{25}{7}}\\ =\dfrac{7}{25}\)
To find sin b:
\(\sin b\\ =\sin(\arctan(\tan b))\\ =\dfrac{\tan b}{\sqrt{1+\tan^2 b}}\\ =\dfrac{-\frac{24}{7}}{\sqrt{1+\dfrac{576}{49}}}\\ =\dfrac{-\frac{24}{7}}{\sqrt{\dfrac{625}{49}}}\\ =\dfrac{-\frac{24}{7}}{\frac{25}{7}}\\ =-\dfrac{24}{25}\)
cos a = 20/29, cos b = 7/25, sin b = -24/25, sin a = -21/29
To find cos(a + b):
\(\cos(a+b)\\ =\cos a\cos b - \sin a \sin b\\ = \left(\dfrac{20}{29}\right)\left(\dfrac{7}{25}\right)-\left(-\dfrac{21}{29}\right)\left(-\dfrac{24}{25}\right)\\ = \left(\dfrac{140}{725}\right) - \left(\dfrac{504}{725}\right)\\ =-\dfrac{364}{725}\)
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