MHB How Can These Trigonometric Equations Be Solved?

[matyw]

Hi all

Can someone answer these equations??

1. find the value of sin(A+B) when sin A = 1/2 and cos B = 3/8

2. solve the equation: 2 - 2 sinx = cos²X when -180° < X < 180°

3. solve rquation for X: 2 tan°X + 2 sec°X - sec X = 12

any help would be great
Cheers
Mat

 

[Greg]

Hello matyw and welcome to MHB! :D

As these are trigonometry questions I've moved your thread into the Trigonometry forum.

Also, we ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[matyw]

Hello matyw and welcome to MHB! :D

As these are trigonometry questions I've moved your thread into the Trigonometry forum.

Also, we ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

Hi Greg

I actually don't know where to start hence why i have posted the equations in full.
My thoughts were to see the answers in full working order, so i could work backwards to see the processes people take to come up with a result. It is hard to be teaching yourself especially when you have no idea of what the results are meant to be
My apologies, i do understand if i don't get help with this.

Regards
Mat

 

[Greg]

These questions assume some background knowledge of trigonometry, specifically (for question 1)

$$\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)$$

and

$$\sin(B)=\pm\sqrt{1-\cos^2(B)},\quad\cos(A)=\pm\sqrt{1-\sin^2(A)}$$

Working with angles in the first quadrant (do you know what the first quadrant is?), can you now solve question 1?

It's somewhat difficult to post good help without knowing what trigonometry you are familiar with. Can you post a brief summary of your trigonometry knowledge?

 

[Greg]

It's been a while and the OP has not responded so I'll post some work.

1. Find the value of sin(A+B) when sin A = 1/2 and cos B = 3/8.

Assuming $A$ and $B$ are in quadrant I,

$$\sin(B)=\sqrt{1-\dfrac{9}{64}}=\dfrac{\sqrt{55}}{8}\quad\cos(A)=\sqrt{1-\dfrac14}=\dfrac{\sqrt3}{2}$$

$$\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)$$

$$=\dfrac12\cdot\dfrac38+\dfrac{\sqrt{55}}{8}\cdot\dfrac{\sqrt3}{2}=\dfrac{3}{16}+\dfrac{\sqrt{165}}{16}=\dfrac{1}{16}\left(3+\sqrt{165}\right)$$

2. Solve the equation 2 - 2 sinx = cos²X when -180° < X < 180°

$$2-2\sin(x)=\cos^2(x)$$

$$2-2\sin(x)=1-\sin^2(x)$$

$$\sin^2(x)-2\sin(x)+1=0$$

$$(\sin(x)-1)^2=0$$

$$\Rightarrow\sin(x)=1,\quad x=90^\circ$$

3. solve rquation for X: 2 tan°X + 2 sec°X - sec X = 12

I'm going to assume this is $2\tan^2(x)+2\sec^2(x)-\sec(x)=12$:

$$2\tan^2(x)+2\sec^2(x)-\sec(x)=12$$

$$2(\sec^2(x)-1)+2\sec^2(x)-\sec(x)-12=0$$

$$4\sec^2(x)-\sec(x)-14=0$$

$$(4\sec(x)+7)(\sec(x)-2)=0$$

$$x=\cos^{-1}\left(-\dfrac47\right),\quad x=\cos^{-1}\left(\dfrac12\right)$$