Given that tan B= 7/24 with B in quadrant III, find sin 2B, cos 2B, and tan 2B, using angle identities.
Did you recognize the 7-24-25 right-angled triangle?
then sin B = -7/25
cos B = -24/25
sin 2B = 2sinBcosB = 2(-7/25)(-24/25) = 336/625
cos2B = cos^2B - sin^2B = 576/625- 49/625 = 527/625
tan2B = sin2B/cos2B = (336/625)÷(527/625) = 336/527
To find sin 2B, cos 2B, and tan 2B, we can use the double-angle formulas.
1. sin 2B = 2 * sin B * cos B
2. cos 2B = cos^2(B) - sin^2(B)
3. tan 2B = (2 * tan B) / (1 - tan^2(B))
Let's calculate the values step-by-step.
Given: tan B = 7/24 and B is in quadrant III.
First, we need to find the values of sin B and cos B.
Using the Pythagorean identity, sin^2(B) + cos^2(B) = 1, we can find sin B and cos B.
1. Since B is in quadrant III, sin B = -sqrt(1 - cos^2(B)).
2. Let's denote cos B as x and substitute it into the equation.
sin B = -sqrt(1 - x^2)
We know that tan B = sin B / cos B.
tan B = (sin B) / x
7/24 = (-sqrt(1 - x^2)) / x
Squaring both sides:
(7/24)^2 = (sqrt(1 - x^2))^2 / (x^2)
49/576 = (1 - x^2) / (x^2)
Rearranging the equation:
49x^2 = 576 - 576x^2
625x^2 = 576
x^2 = 576/625
x = sqrt(576/625)
x = -24/25 (negative value in quadrant III).
Now, we can find sin B:
sin B = -sqrt(1 - (-24/25)^2)
sin B = -sqrt(1 - 576/625)
sin B = -sqrt((625 - 576)/625)
sin B = -sqrt(49/625)
sin B = -7/25
Now that we have sin B and cos B, we can calculate sin 2B, cos 2B, and tan 2B.
1. sin 2B = 2 * sin B * cos B
sin 2B = 2 * (-7/25) * (-24/25)
sin 2B = 336/625
2. cos 2B = cos^2(B) - sin^2(B)
cos 2B = (-24/25)^2 - (-7/25)^2
cos 2B = 576/625 - 49/625
cos 2B = 527/625
3. tan 2B = (2 * tan B) / (1 - tan^2(B))
tan 2B = (2 * (7/24)) / (1 - (7/24)^2)
tan 2B = 14/24 / (1 - 49/576)
tan 2B = 14/24 / (1 - 49/576)
tan 2B = 14/24 / (527/576)
tan 2B = (14/24) * (576/527)
tan 2B = 12/17
Therefore, sin 2B is 336/625, cos 2B is 527/625, and tan 2B is 12/17.
To find the values of sin 2B, cos 2B, and tan 2B, we need to use the angle addition and double-angle identities.
Step 1: Find sin B and cos B
Given that tan B = 7/24 and B is in quadrant III, we can solve for sin B and cos B using the Pythagorean identity:
sin^2 B + cos^2 B = 1
First, let's find cos B.
Since tan B = 7/24, we know that:
tan B = sin B / cos B
7/24 = sin B / cos B
Rearranging the equation, we have:
sin B = (7/24) cos B
Now, let's use the Pythagorean identity to solve for cos B:
(sin B)^2 + (cos B)^2 = 1
[(7/24)cos B]^2 + (cos B)^2 = 1
Expanding and simplifying, we have:
49/576 (cos B)^2 + (cos B)^2 = 1
Combining like terms, we get:
(49/576 + 1) (cos B)^2 = 1
(625/576) (cos B)^2 = 1
Dividing by 625/576, we find:
(cos B)^2 = 576/625
Taking the square root of both sides, we get:
cos B = ± √(576/625)
cos B = ± 24/25
Since B is in quadrant III (where cosine is negative), we take the negative value:
cos B = -24/25
Now, substitute this value into the equation we found earlier:
sin B = (7/24) cos B
sin B = (7/24) (-24/25)
sin B = -7/25
So, we have found that sin B = -7/25 and cos B = -24/25.
Step 2: Use the double-angle identities
Now that we have sin B and cos B, we can use the double-angle identities to find sin 2B, cos 2B, and tan 2B.
sin 2B = 2sin B cos B
cos 2B = cos^2 B - sin^2 B
tan 2B = (2tan B) / (1 - tan^2 B)
Let's calculate these values.
First, sin 2B:
sin 2B = 2sin B cos B
sin 2B = 2(-7/25)(-24/25)
sin 2B = 336/625
Next, cos 2B:
cos 2B = cos^2 B - sin^2 B
cos 2B = (-24/25)^2 - (-7/25)^2
cos 2B = 576/625 - 49/625
cos 2B = 527/625
Finally, tan 2B:
tan 2B = (2tan B) / (1 - tan^2 B)
tan 2B = (2(7/24)) / (1 - (7/24)^2)
tan 2B = 14/24 / (1 - 49/576)
tan 2B = 14/24 / (1 - 49/576)
tan 2B = 14/24 / (527/576)
tan 2B = 1008/1584
tan 2B = 21/33
tan 2B = 7/11
Therefore, sin 2B = 336/625, cos 2B = 527/625, and tan 2B = 7/11.