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
\(\large\sin2\theta \equiv 2\sin \theta \cos \theta \)
\(\large{\cos2\theta \equiv \cos^2\theta -\sin^2\theta\\ \cos2\theta \equiv 2\cos^2\theta -1\\ \cos2\theta \equiv 1 -2\sin^2\theta}\)
This quiz is about the Double Angle formulae for sin2x and cos2x
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
\(cos2x \equiv cos²x - sin²x\)
\(cos2x \equiv 2cos²x – 1\)
Check
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
Check
Given that cosx = 0.4, find the exact value of cos 2x
cos2x \(\equiv \) 2cos²x - 1
cos2x = 2x0.4² - 1
= 2x0.16 - 1
= 0.32 - 1
= - 0.68
Check
Given that sinx = \(\frac{\sqrt{5}}{4}\) , find the exact value of cos 2x
cos2x \(\equiv \) 1 - 2sin²x
cos2x = 1 - 2 \((\frac{\sqrt{5}}{4})^2\)
=1 - 2 \(\frac{5}{16}\)
= 1 - \(\frac{5}{8}\)
\(=-\frac{3}{8}=-0.375\)
Check
Given that cos2x = \(-\frac{7}{8}\) , find the exact value of cosx, given that 0 < x < \(\frac{\pi}{2}\)
cos2x \(\equiv \) 2cos²x - 1
\(-\frac{7}{8}\) = 2cos²x - 1
\(\frac{1}{8}\) = 2cos²x
\(\frac{1}{16}\) = cos²x
cosx = \(\frac{1}{4}\) = 0.25
Check
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 = \(\sqrt{\frac{1-a}{2}}\)
Check
If sin x = \(\frac{12}{13}\) and 0 x < \(\frac{\pi}{2}\)
then sin2x = \(\frac{a}{169}\) . Find a
cosx = \(\frac{5}{13}\)
sin2x = 2sinxcosx
= \(2 \times \frac{12}{13}\times\frac{5}{13}\)
= \(\frac{120}{169}\)
Check
Given that cosx = \(\frac{a}{b}\) , work out cos2x
cos2x \(\equiv \) 2cos²x - 1
cos2x = \(2(\frac{a}{b})^2-1\)
= \(\frac{2a^2}{b^2}-1\)
= \(\frac{2a^2-b^2}{b^2}\)
Check
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 \(\equiv \sqrt{\frac{1-cos2x}{2}}\)
Check
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 = \(2\frac{a}{b}\frac{\sqrt{b^2-a^2}}{b}\)
= \(\frac{2a\sqrt{b^2-a^2}}{b^2}\)
Check
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 a) Expand the brackets
Consider the following identities \(cos^{ 2 }\theta +sin^{ 2 }\theta \equiv 1\\ sin2\theta \equiv 2sin\theta cos\theta \)
b) In order to sketch this function, consider the transformations from the graph of y = sinx
Full Solution
Solve \(cos2θ=sinθ\) for \(0\le \theta \le 2\pi \)
Hint Use the identity \(cos2\theta \equiv 1 -2sin^2\theta\) 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 a) Use the following identity \(cos2\theta \equiv 2cos^2\theta -1\) Full Solution
Let \(cos\theta=\frac{2}{3}\) , where \(0\le \theta \le \frac { \pi }{ 2 } \)
Find the value of
a) \(sin\theta\)
b) \(sin2\theta\)
c) \(sin4\theta\)
Hint a) Draw a triangle and work out \(sin\theta\)
b) Use the identity \(sin2\theta \equiv 2sin\theta cos\theta \)
c) Work out \(cos2\theta\) first. \(2\theta\) is obtuse
Full Solution MY PROGRESS
Self-assessment How much of Double Angle Formulae SL have you understood?
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