In this page, we will will learn about the Double Angle Formulae used in Trigonometry. It is actually quite rare that exam questions are solely about these identities, but it is essential that you can use and manipulate them confidently because they are used in so many different parts of the course (so they do come up a lot!). You will learn what they are and how to use them.
On this page, you should learn about the double angle identities for sine and cosine
\(\large\sin2\theta \equiv 2\sin \theta \cos \theta \)
\(\large{\cos2\theta \equiv \cos^2\theta -\sin^2\theta\\ \cos2\theta \equiv 2\cos^2\theta -1\\ \cos2\theta \equiv 1 -2\sin^2\theta}\)
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START QUIZ!
Consider the trig identities, which of the following is true?
Select all the correct answers.
The only 2 statements that are true come from the double angle formula for cos2x
Consider the trig identities, which of the following is true?
Select all the correct answers.
The only 2 statements that are true come from the double angle formula for sin2x
sin2x \(\equiv \) 2sinxcosx
Given that cosx = 0.4, find the exact value of cos 2x
cos2x =
cos2x \(\equiv \) 2cos²x - 1
cos2x = 2x0.4² - 1
= 2x0.16 - 1
= 0.32 - 1
= - 0.68
Given that sinx = \(\frac{\sqrt{5}}{4}\), find the exact value of cos 2x
cos2x =
cos2x \(\equiv \) 1 - 2sin²x
cos2x = 1 - 2 \((\frac{\sqrt{5}}{4})^2\)
=1 - 2 \(\frac{5}{16}\)
= 1 - \(\frac{5}{8}\)
\(=-\frac{3}{8}=-0.375\)
Given that cos2x = \(-\frac{7}{8}\) , find the exact value of cosx, given that 0 < x < \(\frac{\pi}{2}\)
cosx =
cos2x \(\equiv \) 2cos²x - 1
\(-\frac{7}{8}\) = 2cos²x - 1
\(\frac{1}{8}\)= 2cos²x
\(\frac{1}{16}\)= cos²x
cosx = \(\frac{1}{4}\) = 0.25
Given that cos2x = a and a < 0, find the exact value of sinx
If a < 0 , then cos2x < 0
Therefore 90° < 2x < 270°
and so 45° < x < 135°
Hence 0 < sinx < 1
cos2x \(\equiv \) 1 - 2sin²x
a = 1 - 2sin²x
2sin²x = 1 - a
sin²x = \(\frac{1-a}{2}\)
Since sinx is positve, we take the positive root
sinx = \(\sqrt{\frac{1-a}{2}}\)
If sin x = \(\frac{12}{13}\) and 0 x < \(\frac{\pi}{2}\)
then sin2x = \(\frac{a}{169}\). Find a
a =
cosx = \(\frac{5}{13}\)
sin2x = 2sinxcosx
= \(2 \times \frac{12}{13}\times\frac{5}{13}\)
= \(\frac{120}{169}\)
Given that cosx = \(\frac{a}{b}\) , work out cos2x
cos2x \(\equiv \) 2cos²x - 1
cos2x = \(2(\frac{a}{b})^2-1\)
= \(\frac{2a^2}{b^2}-1\)
= \(\frac{2a^2-b^2}{b^2}\)
Which of the following is true?
The two correct answers are from rearranging the double angle formula for cos2x
cos2x \(\equiv \) 2cos²x - 1
cos2x + 1 \(\equiv \) 2 cos²x
\(\frac{cos2x+1}{2} \equiv\) cos²x
\(\sqrt{\frac{cos2x+1}{2} }\equiv\) cosx
cos2x \(\equiv \) 1 - 2sin²x
2sin²x \(\equiv\) 1 - cos2x
sin²x \(\equiv \frac{1-cos2x}{2}\)
sinx \(\equiv \sqrt{\frac{1-cos2x}{2}}\)
Given that sinx = \(\frac{a}{b}\), work out sin2x
If sinx = \(\frac{a}{b}\), then cosx = \(\frac{\sqrt{b^2-a^2}}{b}\)
sin2x \(\equiv\)2sinxcosx
sin2x = \(2\frac{a}{b}\frac{\sqrt{b^2-a^2}}{b}\)
= \(\frac{2a\sqrt{b^2-a^2}}{b^2}\)
Let f(x) = (cos2x - sin2x)²
a) Show that f(x) can be expressed as 1 - sin4x
b) Let f(x) = 1 - sin4x. Sketch the graph of f for \(0\le x\le \pi \)
Hint
Full Solution
a) Show that \(cos2\theta-3cos\theta+2\equiv 2{ cos }^{ 2 }\theta -3cos\theta +1\)
b) Hence, solve \(cos2\theta-3cos\theta+2=0\) for \(0\le \theta \le 2\pi \)
Hint
Full Solution
How much of Double Angle Formulae SL have you understood?
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