Trigonometrical Identities and simple equations!

Can someone pls help me on this question:

EX 10 A, C 2:

Find, without using your calculator the value of:
sin(theta) and cos(theta), given tht tan(theta) = 5/12 and (theta) is acute

Thanks

Reply 1

beast

hey there.

Right, we know that tan (theta) = 5/12.

now use the identity, tan(theta) = opposite/adjacent.

Hence , the side opposite theta is 5.
the side adjacent to theta is 12.

Therefore, the hypotenuse = the square root of (5^2 + 12^2) = 13.

Hence, sin(theta) = sin(theta) = opposite/hypotenuse = 5/13
And cos(theta) = adjacent/hypotenuse = 12/13

Reply 2

Mohit_C

Cheers m8, thanks 4 the help

Reply 3

Mohit_C

Ok, I got another question.

In each of the following, eliminate (theta) to give an equation relating x and y:

a) x = sin (theta), y = cos (theta)

b) x = sin (theta), y = 2cos(theta)

c) x = sin (theta), y = cos^2 (theta)

d) x = sin (theta), y = tan (theta)

e) x = sin (theta) + cos (theta), y = cos (theta) - sin (theta)

Thanks.

Reply 4

dvs

For the first 3 use the identity sin^2(t) + cos^2(t) = 1.

For d you could write y = sin(t)/cos(t) = sin(t)/sqrt(1 - sin^2(t)) = x/sqrt(1 - x^2).

And for e note that (x + y)/2 = cos(t) and (x - y)/2 = sin(t).

Reply 5

Mohit_C

Can you pls explain how you got tht? I dont understand this.

Reply 6

yusufu

In each of the following, eliminate (theta) to give an equation relating x and y:

a) x = sin (theta), y = cos (theta)

x² = sin²t , y² = cos²t
x² + y² = 1

b) x = sin (theta), y = 2cos(theta)

x² = sin²t , (y/2)² = cos²t
x² + y²/4 = 1

c) x = sin (theta), y = cos^2 (theta)

x² = sin²t
x² + y =1

d) x = sin (theta), y = tan (theta)

x² = sin²t , y² = sin²t / cos²t
cos²t = 1-sin²t

y² = x² / (1-x²)

e) x = sin (theta) + cos (theta), y = cos (theta) - sin (theta)
x + y = 2cost , x - y = 2sint
((x+y)/2)² = cos²t , ((x-y)/2)² = sin²t

(x+y)² + (x-y)² = 4
x² + 2xy +y² + x² - 2xy + y² = 4
2x² + 2y² = 4

Reply 7

Mohit_C

Another queston:

Ex 10B, 2d)
Solve the following equations for x: cosxº: 0.84, -360<x<0

3d) sin (theta) = tan (theta), 0<(theta)<(or equal to)2(pi)

e) 2(1+tan(theta))= 1-5 tan (theta), -(pi)<(theta)<(or equal to)2(pi)

Thanks.

PS. is there a way of putting in pi, theta and more/less than or equal to signs? makes things easier.

Reply 8

welwyn

3d) use tan = sin/cos and times up to get

sinxcosx = sinx - then factorise (don't divide by sinx)

to get sinx(cosx - 1) = 0, so either cosx = 1 or sinx = 0.

then solve (use Cast diag)

3e) 2 + 2tanx = 1 - 5tanx

7tanx = -1

tanx = -1/7 >>> solve (remember sin^2(x) + cos^2(x) = 1)

Reply 9

Mohit_C

I'm really getting stuck. When the interval is between -245<x<565 and then you have an angle at -27.3 and one at 152º then how do you get the values for X. Do you go first -245º then go positively until 565 (i.e. anti-clockwise) or do you just count how many degrees till the angle's -27.3º and 152º??

PLS HELP!

Reply 10

Mohit_C

Can someone pls explain how to solve this: 3tanx=2tan^2(x)

Thanks

Reply 11

sahra-t

It's just a quadratic. Put tan x= y, so it becomes

3y = 2y²
2y² - 3y = 0
y (2y - 3) = 0

Solve for y. Then once you've got that, y = tan x, and work out the values of x.

Ok, thanks

Ok, I got these last two questions.

1) Ex 10E. 9b) Solve 4cos3xº-sin(90-3x)º=2 in the interval 0 ≤ x ≤ 360. In 9a) you had to find a single trigonometric function of "4cos3xº-sin(90-3x)º" and I got that it gives 3cos3x. But when I checked my answer to 9b) i saw that i missed two solutions.

2) 19a) [There's a diagram of the cos graph but it is moved to the rite and has three intersections A, B and C with x-axis]
The diagram shows part of the curve with equation Y = cos (px-q)º, where p and q are positive constants and q < 180. The curve cuts the x-axis at points A, B and C.

Given that coordinates of A and B are (100,0) and (220,0) repectively:
i) Write down the coordinates of C.
ii) Find the value of p and the value of q.

Thanks.