\text { Solve } 1-\cos \left(\frac{x}{2}\right)=\cos \left(\frac{x}{2}\right) \text { for } x \in[-2 \pi, \pi]

If you answer “no”, it is now time to improve your Maths Methods knowledge and skills up to the VCE level.


where to start??

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Still hesitated? No worries. Let’s take a look at the solution for the above trigonometric question first

Gives \cos \left(\frac{x}{2}\right)=\frac{1}{2}.

From special values gives \frac{x}{2}=\frac{\pi}{3}.

\cos \left(-\frac{x}{2}\right)=\frac{1}{2}, so -\frac{x}{2}=\frac{\pi}{3}.

Gives \frac{x}{2}=\frac{\pi}{3}+2 n \pi, x=\frac{2 \pi}{3}+4 n \pi. The other one gives x=-\frac{2 \pi}{3}+4 n \pi

We have: x=\pm \frac{2 \pi}{3}+4 n \pi, n \in Z.

Which is x=-\frac{2 \pi}{3}, \frac{2 \pi}{3}.

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For the above cosine question, students can easily find trigonometry related knowledge and tutorials required by VCE in A+ MM Master Notes. Take a quick glimpse at some examples.

Trigonometry Values Table
Differentiation Rules for Trigonometric Functions
QnA Tutorials for Trigonometric Functions


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