MHB How Can These Trigonometric Equations Be Solved?

  • Thread starter Thread starter matyw
  • Start date Start date
AI Thread Summary
The discussion revolves around solving three trigonometric equations. The first equation involves finding sin(A+B) using given values for sin A and cos B, with the solution provided showing the use of trigonometric identities. The second equation simplifies to find sin(x) = 1, leading to the solution x = 90°. The third equation is solved by assuming a specific form and applying algebraic manipulation to find solutions for x. The importance of understanding trigonometric identities and providing prior work for assistance is emphasized throughout the discussion.
matyw
Messages
2
Reaction score
0
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
 
Mathematics news on Phys.org
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?
 
greg1313 said:
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
 
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?
 
It's been a while and the OP has not responded so I'll post some work.

matyw said:
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,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$x=\cos^{-1}\left(-\dfrac47\right),\quad x=\cos^{-1}\left(\dfrac12\right)$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top