# Can anyone help w/ these.

1) Solve the equation in the internal [0deg, 360deg].

a) sin 2x = -sin x

b) sin 2T = -1/2 (where T is angle)

c) 4 sin^2T = 3

2) Evaluate the expression.

sin(arctan 2)

3) Rewrite the following w/o using trigonometric or inverse tirgonometric functions.

cos(arcsin u)

In a use the double angle formula: Sin2X= 2sinXcosX

When you divide through by sinX, be certain to remember to test to see if X=0 is a solution also.

In b, solve for 2T (then divide it by two).

in C, take the square root of each side, solve for sinT, and then T.

in 2, remember solve for arcTan2, then what is the sine of that?

In 3, remember that if sine = u, then cos of the complementary angle is u.

8 months ago

## 1) Solve the equation in the interval [0Â°, 360Â°]:

a) sin 2x = -sin x:

To solve this equation, we can use the double angle formula for sine, which states that sin 2x = 2sin x cos x.

So, we have 2sin x cos x = -sin x.

Dividing both sides of the equation by sin x, we get:

2cos x = -1.

Now, we need to solve for x. To do that, let's consider the possible values for cos x when it is equal to -1/2.

Using the unit circle or the trigonometric ratios, we find that cos x = -1/2 when x = 120Â° and x = 240Â°.

However, we have to check the validity of these solutions by substituting them back into the original equation. We see that x = 120Â° satisfies the equation, while x = 240Â° does not.

Therefore, the solution to the equation sin 2x = -sin x in the interval [0Â°, 360Â°] is x = 120Â°.

b) sin 2T = -1/2:

Similar to part a, we can use the double angle formula for sine to rewrite the equation as 2sin T cos T = -1/2.

Dividing both sides by 2 and rearranging, we get:

sin T cos T = -1/4.

Now, we need to solve for 2T. We can find the value for 2T that satisfies this equation by considering the values of sin T and cos T when their product is equal to -1/4.

Using the unit circle or the trigonometric ratios, we find that sin T = -1/2 and cos T = -1/2 satisfy the equation.

Now, solve for 2T:

2T = sin^-1(-1/2) = 210Â° or 330Â°.

Finally, divide 2T by 2 to obtain the solutions for T: T = 105Â° or 165Â°.

Therefore, the solutions to the equation sin 2T = -1/2 in the interval [0Â°, 360Â°] are T = 105Â° and T = 165Â°.

c) 4 sin^2T = 3:

First, let's isolate sin^2T by dividing both sides of the equation by 4:

sin^2T = 3/4.

Next, take the square root of both sides to find sin T:

sin T = Â±âˆš(3/4) = Â±âˆš3/2.

Since sin T can be positive or negative, we need to consider both possibilities.

Using the unit circle or the trigonometric ratios, we find that sin T = âˆš3/2 when T = 60Â° or 300Â°.

Similarly, sin T = -âˆš3/2 when T = 120Â° or 240Â°.

Therefore, the solutions to the equation 4 sin^2T = 3 in the interval [0Â°, 360Â°] are T = 60Â°, T = 120Â°, T = 240Â°, and T = 300Â°.

2) Evaluate the expression:

sin(arctan 2):

To evaluate this expression, first find the value of arctan 2. This can be done using a calculator or by finding the angle whose tangent is 2. The value of arctan 2 is approximately 63.43Â°.

Now, we need to find the sine of 63.43Â°. Using either a calculator or the unit circle, we find that sin(63.43Â°) is approximately 0.896.

Therefore, sin(arctan 2) â‰ˆ 0.896.

3) Rewrite the following without using trigonometric or inverse trigonometric functions:

cos(arcsin u):

When arcsin u = Î¸, it means that sin Î¸ = u.

Let's assume that sin Î¸ = u and cos Î¸ = v.

Now, using basic trigonometric properties:

sin^2 Î¸ + cos^2 Î¸ = 1.

Substitute the values of sin Î¸ and cos Î¸:

u^2 + v^2 = 1.

Since sin Î¸ = u, we have:
v = Â±âˆš(1 - u^2).

Therefore, cos(arcsin u) can be rewritten as Â±âˆš(1 - u^2) depending on the positive or negative value of v.

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