11. (a) Show that
$$\sin 2x \tan x = 1 - \cos 2x$$, where
$$\frac{\pi}{2} < x < \frac{3\pi}{2}$$ - Scottish Highers Maths - Question 11 - 2016
Question 11
11. (a) Show that
$$\sin 2x \tan x = 1 - \cos 2x$$, where
$$\frac{\pi}{2} < x < \frac{3\pi}{2}$$.
(b) Given that
$$f(x) = \sin 2x \tan x$$, find
$$f... show full transcript
Worked Solution & Example Answer:11. (a) Show that
$$\sin 2x \tan x = 1 - \cos 2x$$, where
$$\frac{\pi}{2} < x < \frac{3\pi}{2}$$ - Scottish Highers Maths - Question 11 - 2016
Step 1
Show that
$$\sin 2x \tan x = 1 - \cos 2x$$
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Answer
To prove the identity, start with the left-hand side (LHS):
Substitute the definitions of sine and tangent:
sin2xtanx=sin2x⋅cosxsinx.
Using the double-angle formula for sine, we have:
sin2x=2sinxcosx.
Therefore,
sin2xtanx=2sinxcosx⋅cosxsinx=2sin2x.
Now, consider the right-hand side (RHS):
Using the double angle formula for cosine, we know:
cos2x=1−2sin2x.
Rearranging gives:
1−cos2x=1−(1−2sin2x)=2sin2x.
Thus, LHS = RHS:
sin2xtanx=1−cos2x.
Therefore, the identity is shown.
Step 2
Given that
$$f(x) = \sin 2x \tan x$$, find
$$f'(x)$$.
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Answer
To find the derivative of the function:
Use the product rule, where if you have
u=sin2x and
v=tanx:
u′=2cos2x
and
v′=sec2x.
Then, by the product rule:
f′(x)=u′v+uv′.
Substituting in:
f′(x)=(2cos2x)(tanx)+(sin2x)(sec2x).
Therefore, the derivative is:
f′(x)=2cos2xtanx+sin2xsec2x.
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