how do you establish the identity of cos(3pi/2+0)=sin0
is the "0" in the problem, the angle theta? if it is,
cos(3pi/2+[theta]) = sin [theta]
*expand cos(3pi/2+[theta]),
cos(3pi/2)cos(theta) - sin(3pi/2)sin(theta) = sin [theta]
*cos(3pi/2) = 0 and sin(3pi/2) = -1, thus:
-(-sin [theta]) = sin [theta]
sin [theta] = sin [theta]
so there,, :)
To establish the identity cos(3π/2 + 0) = sin(0), we'll break it down step-by-step using trigonometric identities.
Step 1: Simplify the angle within the cosine function
In this case, the angle is (3π/2 + 0). The given zero term doesn't affect the equation, so we can simplify it to cos(3π/2).
Step 2: Identify the value of cos(3π/2)
Using the unit circle or knowing the properties of trigonometric functions, we can determine that cos(3π/2) is equal to 0.
Step 3: Rewrite the equation
We now have cos(3π/2) = sin(0).
Step 4: Determine the value of sin(0)
The sine of 0 degrees is 0.
Step 5: Expand the equation
We now have 0 = 0.
Step 6: Analyze the equation
Since both sides of the equation are equal to 0, we can conclude that the identity cos(3π/2) = sin(0) holds true.
Therefore, we have established the identity cos(3π/2 + 0) = sin(0).
To establish the identity of cos(3π/2 + 0) = sin(0), we will break down the problem step by step.
First, let's evaluate cos(3π/2) and sin(0) separately:
cos(3π/2) refers to the cosine function evaluated at 3π/2. Let's recall the unit circle as a reference:
```
π/2
|
|
______|_______ (x = 0, y = 1)
|
π ------------|------------- 2π
|
|
3π/2
```
Looking at the unit circle, we can see that at 3π/2, the x-coordinate is 0, and the y-coordinate is -1. Therefore, cos(3π/2) = 0.
Next, sin(0) refers to the sine function evaluated at 0. At 0, the x-coordinate is 1, and the y-coordinate is 0. Therefore, sin(0) = 0.
Now let's compare the two sides of the given equation:
cos(3π/2 + 0) = sin(0)
We already determined that cos(3π/2) = 0 and sin(0) = 0, so substituting these values, we have:
0 = 0
Since 0 is equal to 0, the given equation holds true. Therefore, the identity cos(3π/2 + 0) = sin(0) is established.
To me this appears to be a silly question, if your 0 is indeed a zero.
Why even write it as (3π/2 + 0)
it would simply be cos 3π/2
and you should know that cos 3π/2 = 0 which is also the value of the right side.
End of problem.