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secondary 3 | A Maths
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idk how to do d) please help asap
Date Posted: 2 years ago
Views: 229
You got cos A as 4/5, which is correct.
Using the double angle formula,
cos A = 2 cos^2 (A/2) - 1
[Yes, the angle works for A vs A/2 since A is double of A/2, it’s not just limited to 2A vs A]
4/5 = 2 cos^2 (A/2) - 1
9/5 = 2 cos^2 (A/2)
9/10 = cos^2 (A/2)
Taking square roots,
cos (A/2) = 3 / sqrt 10 or -3 / sqrt 10
However, we can only pick one of them.
Notice that 270 < A < 360 (in degrees). Halve this range and we get 135 < A/2 < 180, which is well within the second quadrant. As we know it, the cosine of an angle in the second quadrant has a negative output.
So, cos (A/2)
= -3 / sqrt 10
= - (3 sqrt 10) / 10
Using the double angle formula,
cos A = 2 cos^2 (A/2) - 1
[Yes, the angle works for A vs A/2 since A is double of A/2, it’s not just limited to 2A vs A]
4/5 = 2 cos^2 (A/2) - 1
9/5 = 2 cos^2 (A/2)
9/10 = cos^2 (A/2)
Taking square roots,
cos (A/2) = 3 / sqrt 10 or -3 / sqrt 10
However, we can only pick one of them.
Notice that 270 < A < 360 (in degrees). Halve this range and we get 135 < A/2 < 180, which is well within the second quadrant. As we know it, the cosine of an angle in the second quadrant has a negative output.
So, cos (A/2)
= -3 / sqrt 10
= - (3 sqrt 10) / 10
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Hope this helps
Date Posted:
2 years ago
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