Math Help - Trigonometry

1. Trigonometry

.)Solve the equation sin(x+170) cos(x-120) =0.7, for 0≤x≤3600
2.) Prove that sin 3x+sin2x 3x+2x/2 cos3x-2x/2
3.) Find all values x such that 0≤x≤3600 by solving cos2 x+sinx-3=0

4.) If sina=3/5 and cosb=9/41, find the values of sin (a-b),aand b being acute angle

2. Originally Posted by PRINCE MARVELLOUS
4.) If sina=3/5 and cosb=9/41, find the values of sin (a-b),aand b being acute angle
First, we use the sine of a difference identity:
$sin(a-b) = sin(a)cos(b) - cos(a)sin(b)$
We know sin(a) & cos(b), but we don't know cos(a) or sin(b). Or do we?

We can figure it out by drawing two right triangles. On the first triangle, label one of the acute angles as "a." sine is defined as opposite/hypotenuse, so label the opposite side 3 and the hypotenuse 5. Use the Pythagorean theorem to find the missing side (adjacent). Then find cos(b) (adjacent over hypotenuse).

Do it again with another triangle. Label one of the acute angles as "b." cos(b) = 9/41, so the adjacent side is 9 and the hypotenuse is 41. Use the Pythagorean theorem again to find the missing side (opposite). Now find sin(b).

Finally, plug in all the ratios into the identity I wrote above and evaluate.

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3. Appreciation

l thank you for putting me through in one of my questions,but after l find the missing side of sin a and that of missing side of sin b in which l plug into the identity you gave me,l still not getting it, please more explaination and looking forward to hear from the remaining post or question.

Thanks

4. Okay, first the triangle with angle a. The hypotenuse is 5, and the opposite is 3. Use the Pythagorean Theorem to find the remaining side:
$c^{2} = a^{2} + b^{2}$ (note: a and b here are not the same as the angles a and b mentioned above. c is the hypotenuse.)
$5^{2} = a^{2} + 3^{2}$
$25 = a^{2} + 9$
$16 = a^{2}$, so the remaining side (adjacent) is 4.
$cos(a) = \frac{adj}{hyp} = \frac{4}{5}$.

Now you try with the other triangle, the one with angle b.

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5. Explanation

First the triangle with angle a .Two of the sides are 3 &5,and l got the length of the remaining side to be 4.i.e as you get your own,while the second triangle with angle b .Two of the sides are 9 & 41 ,l got the length of the remaining to side to be 40.Then l finally equated it as sin(a-b)=sin(3/5-5/40)=sin(a-b)=3/40 but l dont know lf am right.Please confirm it.And also please out of my four questions which l posted the number 1,question which l put sin(x+17)cos(x-12)=0.7, the figure is not 170 & 120, but x+17degree & x-12degree but am trying to edit it but no way.

Thanks yeongil

6. sin(a-b)=sin(3/5-5/40)=sin(a-b)=3/40
You're confusing sin(a) with a. sin(a) = 3/5, not a = 3/5.

You want to use this formula:
$sin(a-b) = sin(a)cos(b) - cos(a)sin(b)$
Substitute:
$sin(a-b) = \left(\frac{3}{5}\right)\left(\frac{9}{41}\right) - cos(a)sin(b)$
I showed that cos(a) = 4/5:
$sin(a-b) = \left(\frac{3}{5}\right)\left(\frac{9}{41}\right) - \left(\frac{4}{5}\right)sin(b)$

On the 2nd triangle, you found the remaining side to be 40. So what's sin(b)?

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7. Explanation

Thanks for putting me through.This is how l calculated it,
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
sin(3/5)cos(9/41)-cos(4/5)sin(40/41)
(3/5)(9/41)-(4/5)(40/41)

27/205-160/205

27-160/205

=-133/205 that is the answer l get,but please tell me if am right and am looking forward to hear more explanation on the remaining questions

8. Originally Posted by Prince Marvellous
Thanks for putting me through.This is how l calculated it,
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
sin(3/5)cos(9/41)-cos(4/5)sin(40/41) <---DON'T DO THAT!!!!
(3/5)(9/41)-(4/5)(40/41)
Again, watch out! Don't write "sin(3/5)", "cos(9/41)", etc. What goes inside the sine and cosine functions are angles. We don't know what they are at the moment, so we call them a and b. sin(a) = 3/5. sin(3/5) <> 3/5! Cross out the line where I wrote "don't do that" above.

=-133/205 that is the answer l get,but please tell me if am right and am looking forward to hear more explanation on the remaining questions

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9. Appreciation

l really appreciated how you are have be helping me all this while,but you didn't say maybe am right or not and also l never receive any instruction or guideline on the remaining three (3) questions.please am looking forward

10. First off, you when wrote 0 ≤ x ≤ 3600, do you really mean 0 ≤ x ≤360°(degrees)?

Originally Posted by Prince Marvellous
2.) Prove that sin 3x+sin2x 3x+2x/2 cos3x-2x/2
Can you add parentheses so that it's easier to read? And where is the equal sign?

3.) Find all values x such that 0≤x≤3600 by solving cos2 x+sinx-3=0
I assume you mean $cos^{2}x + sin x - 3 = 0$.
Use the Pythagorean identity to change cos^2 (x) in terms of sin^2 (x):
$1 - sin^{2}x + sin x - 3 = 0$
$-sin^{2}x + sin x - 2 = 0$
$sin^{2}x - sin x + 2 = 0$
Factor:
$(sin x + 1)(sin x - 2) = 0$
Now set each factor equal to zero and solve for x.

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11. Thanks

Thanks very much for putting me through,on question,l mean 360degree,thank for that,

And this is what l get,(sinx+1)(sinx-)=0
x+1=0
x-2=0
x=-1
Therefore,since x= -1,it fall within 3rd &4th quadrant where sine is negative
and the values of x is 270degree & 360degree.Please l my right or any addition

And on the question 2, am sorry for the omitted the equal sign,and this is the right one. prove that sin3x+sin2x= 2sin 3x+2x/2cos3x-2x/2

please how can l find icon for thank is not among lists?

12. Originally Posted by Prince Marvellous
Thanks very much for putting me through,on question,l mean 360degree,thank for that,

And this is what l get,(sinx+1)(sinx-)=0
x+1=0
x-2=0
x=-1
Therefore,since x= -1,it fall within 3rd &4th quadrant where sine is negative
and the values of x is 270degree & 360degree.Please l my right or any addition
No, no, no, you really have to be precise in your notation.
(sin x + 1)(sin x - 2) = 0
means that either
sin x + 1 = 0 or sin x - 2 = 0.

sin x + 1 = 0
sin x = -1
x = 270°

sin x - 2 = 0
sin x = 2
There is no solution here, because the sine of any angle is between -1 and 1.

So the only answer is x = 270°.

P.S. Your #2 is still not clear. I don't know what is being divided exactly. Use parentheses "( )".

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13. Appreciation

Thanks very for your assistance all this while,anyway this it ,PROVE THAT (Sin3x+2x)=2Sin(3x+2x)/2Cos(3x-2x)/2: l meany (3x+2x)over 2 & (3x-2x)over 2.PLS am waiting for question no,1

14. Originally Posted by Prince Marvellous

.)Solve the equation sin(x+170) cos(x-120) =0.7, for 0≤x≤3600
[snip]
Start by applying the formula $\sin A \cos B = \frac{\sin (A + B) + \sin (A - B)}{2}$.

15. Explanation

Sin3x+Sin2x=2Sin(3x+2x)/2 Cos(3x-2x)/2
Note:Sinx+Siny=2Sinx+y/2 Cos x-y/2
and also Cosx=Sinx

Sin(3x+2x)+(3x-2x)/2 Sin (3x+2x)-(3x-2x)/2

Sin 5x+x/2 Sin 5x-x/2

Sin6x/2 Sin4x/2
=Sin3x+Sin2x
Therefore Sin3x+2x=2 Sin (3x+2x)/2 Cos (3x-2x)/2

But the second question am still confused the way you said l should do it
l mean this one Solve the equation Sin(x+17degree) Cos (x-12degree)=0.7,for 0 lessthan or equal to x less than or equal to 360 degree.