I have a question!

Given: 

sin a= -21/29 , with 3pie/2 <a <2pie

and

tan B= -24/7 ,with pie/2 <B< pie

Find cos(a+b)

1.

\(\begin{array}{|rcll|} \hline \sin(a) &=& \frac{-21}{29}, \text{ with } \frac{3\pi}{2} <a< 2\pi \quad \text{IV. Quadrant} \\ a &=& \arcsin(\frac{-21}{29}) \\ \mathbf{a} & \mathbf{=} & \mathbf{-0.80978357257\ \text{rad } \quad (-46.3971810273^{\circ}) }\\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \tan(b) &=& \frac{24}{-7}, \text{ with } \frac{\pi}{2} <b< \pi \quad \text{II. Quadrant} \\ b &=& \arctan(\frac{24}{-7}) \\ b &=& -1.28700221759 + \pi \quad (\text{II. Quadrant}) \\ \mathbf{b} & \mathbf{=} & \mathbf{1.85459043600 \ \text{rad } \quad (106.260204708^{\circ}) }\\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline \cos(a+b) &=& \cos(-0.80978357257\ \text{rad}+1.85459043600 \ \text{rad } ) \\ &=& \cos(1.04480686343 \ \text{rad } ) \\ \mathbf{\cos(a+b)} & \mathbf{=} & \mathbf{0.50206896552} \\ \hline \end{array}\)

Given: 

sin a= -21/29 , with 3pie/2 <a <2pie

and

tan B= -24/7 ,with pie/2 <B< pie

Find cos(a+b)

\(\frac{3\pi}{2}<\alpha<2\pi\)      \(4.712..<\alpha<6.2831..\)

\(sin \alpha=-\frac{21}{29}=0.72413..\)  

\(\alpha= arc sin(-\frac{21}{29 })=-0.80978..+2\pi =\color{blue}5.47340..\) 

\(\frac{3\pi}{2}<5.47340<2\pi\) 

\(\frac{3\pi}{2}<\beta<2\pi\)

\(tan\beta=-\frac{24}{7}=-3.42857..\) 

\(\beta=arctan(-\frac{24}{7})=-1.287..+2\pi=\color{blue}4.99618..\) 

\(\frac{3\pi}{2}<4.99618<2\pi\)

\(\Large cos(\alpha+\beta)=cos(arcsin(-\frac{21}{29})+arctan(-\frac{24}{7}))\)

\(cos(\alpha+\beta)=cos(5.47340173461+4.99618308959)\) 

\(\Large cos(\alpha+\beta)=-0.502068965519\)

  !

sin A =  -21/29

cos A  = 20/29

sin B = 24/25

cos B = -7/25

cos ( A + B) =

cos(A)cos(B) - sin(A)sin(B) =

[20/29 * -7/25]   -  [-21/29 * 24/25 ]  =

[ -140 + 504 ] / 725  =

364 / 725  ≈  0.5020689655172414

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):

\(\cos(a+b)\\ =\cos a\cos b - \sin a \sin b\\ = \left(\dfrac{20}{29}\right)\left(\dfrac{7}{25}\right)-\left(-\dfrac{21}{29}\right)\left(-\dfrac{24}{25}\right)\\ = \left(\dfrac{140}{725}\right) - \left(\dfrac{504}{725}\right)\\ =-\dfrac{364}{725}\)