Angle a lies in the second quadrant and sina=7/25.
a) Determine an exact value for cos 2a
b) Determine an exact value for sin 2a
Before you do anything, start by drawing this angle on the unit circle (this does not have to be to scale). Because angle π’ is in quadrant 2, the domain of the angle must be between π and π/2. Your graph should look like a single line between π and π/2. Now, draw a dotted line from the tip of your line pointing to the left side of your graph down to the x-axis. Draw angle π’ between the x-axis and your line. You should now have a triangle in the second quadrant.
Now take the information you have for sine to construct your triangle. We know that to find sine, we divide the opposite side of the triangle by its hypotenuse. So, if sinπ’ = 7/25, then the opposite side of our triangle is 7 and the hypotenuse must be 25. Find the length of the last side by using the pythagorean theorem:
a^2 + b^2 = c^2
b^2 = c^2 - a^2
b = β(25)^2 - (7)^2)
b = β625 - 49
b = β576
b = 24
Therefore, our triangle has a side of 7 opposite to π’, another side of 24 adjacent to π’, and a hypotenuse of 25. Now we can solve for the next questions.
a) Use the triangle and the double angle identity to solve for cos2π’:
(Remember, because we are in quadrant 2, the CAST rule says cos is negative, so we need to indicate that at the start. Sine however is positive, so we can leave that as it is.)
-cos2π’ = -cos^2π’ - sin^2π’
= (-24/25)^2 - (7/25)^2
β Because -24/25 is squared, the negative sign disappears
= (576/625) - (49/625)
= 527/625
Therefore, cos2π’ is 527/625.
b) Use the triangle and the double angle identity to solve for sin2π’:
sin2π’ = 2sinπ’ cosπ’
= 2(7/25)(-24/25)
β Remember cos is negative
= 2(-168/625)
= -336/625
Therefore, sin2π’ is -336/625.
Oh, I have just the right joke for this one. Why did the angle go to the therapist?
Because it had cosine identity crisis!
Now let's solve this identity crisis for angle a in the second quadrant and sine a = 7/25.
a) To find an exact value for cos 2a, we can use the double-angle formula for cosine:
cos 2a = cos^2(a) - sin^2(a)
Since we already know sine a, we can find cosine a using the Pythagorean identity: sin^2(a) + cos^2(a) = 1.
Let's find cosine a first:
cos^2(a) = 1 - sin^2(a)
cos^2(a) = 1 - (7/25)^2
cos^2(a) = 1 - 49/625
cos^2(a) = 576/625
cos(a) = Β±sqrt(576/625)
cos(a) = Β±24/25
Since angle a lies in the second quadrant, the cosine value will be negative. Therefore:
cos(a) = -24/25
Now, let's find cos 2a:
cos 2a = cos^2(a) - sin^2(a)
cos 2a = (-24/25)^2 - (7/25)^2
cos 2a = 576/625 - 49/625
cos 2a = 527/625
So, the exact value for cos 2a is 527/625.
b) To find an exact value for sin 2a, we can use the double-angle formula for sine:
sin 2a = 2sin(a)cos(a)
Plugging in the values we already know:
sin 2a = 2 * (7/25) * (-24/25)
sin 2a = -336/625
Therefore, the exact value for sin 2a is -336/625.
And there you have it! The angle a might have had an identity crisis, but now we know the exact values for cos 2a (-527/625) and sin 2a (-336/625).
To find the exact value of cos 2a and sin 2a, we can make use of the trigonometric identities.
First, let's find the value of cos a using the Pythagorean identity:
cos^2(a) + sin^2(a) = 1
cos^2(a) + (7/25)^2 = 1
cos^2(a) + 49/625 = 1
cos^2(a) = 1 - 49/625
cos^2(a) = (625 - 49)/625
cos^2(a) = 576/625
cos(a) = β(576/625)
cos(a) = 24/25
Since angle a lies in the second quadrant, cos a is negative. Therefore, cos a = -24/25.
Now, we can find the values of cos 2a and sin 2a.
a) Using the double-angle identity for cosine:
cos 2a = cos^2(a) - sin^2(a)
= (-24/25)^2 - (7/25)^2
= (576/625) - (49/625)
= (576 - 49)/625
= 527/625
b) Using the double-angle identity for sine:
sin 2a = 2sin(a)cos(a)
= 2(7/25)(-24/25)
= -48/625
So, the exact values are:
a) cos 2a = 527/625
b) sin 2a = -48/625
To find the exact values for cos 2a and sin 2a, we first need to determine the value of cos a.
To find the value of cos a, we can use the trigonometric identity: cos^2 a + sin^2 a = 1. Given that sin a = 7/25, we can substitute the value into the equation and solve for cos a:
cos^2 a + (7/25)^2 = 1
cos^2 a + 49/625 = 1
cos^2 a = 1 - 49/625
cos^2 a = 576/625
cos a = Β±β(576/625)
cos a = Β±24/25
Since the angle a lies in the second quadrant, the cosine value would be negative. Therefore, we have:
cos a = -24/25
Now that we have the value of cos a, we can find the exact values for cos 2a and sin 2a using the double-angle formulas:
cos 2a = cos^2 a - sin^2 a
sin 2a = 2sin a * cos a
a) Determine an exact value for cos 2a:
Using the value of cos a we found earlier, we substitute it into the formula:
cos 2a = (cos^2 a - sin^2 a)
cos 2a = ((-24/25)^2 - (7/25)^2)
cos 2a = (576/625 - 49/625)
cos 2a = 527/625
So the exact value of cos 2a is 527/625.
b) Determine an exact value for sin 2a:
Using the values of sin a and cos a, we substitute them into the formula:
sin 2a = 2sin a * cos a
sin 2a = 2 * (7/25) * (-24/25)
sin 2a = -336/625
So the exact value of sin 2a is -336/625.
There must be a 7,24,25 triangle involved.
cosa=-24/25
cos2a=sin^2 a - cos^2 a
sin2a=2sinacosa