# How do you use double angle formulas to calculate cos 2x and sin 2x without finding x if #cos x = 3/5# and x is in the first quadrant?

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# cos2x = -7/25 #

# sin2x = 24/25 #

Next we use the sine and cosine double angle formula, so that:

And:

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To calculate (\cos 2x) and (\sin 2x) without finding (x), if (\cos x = \frac{3}{5}) and (x) is in the first quadrant, use the double angle formulas:

[ \cos 2x = 2(\cos^2 x) - 1 ] [ \sin 2x = 2\sin x \cos x ]

Given (\cos x = \frac{3}{5}), first find (\sin x) using the Pythagorean identity:

[ \sin^2 x + \cos^2 x = 1 ] [ \sin^2 x = 1 - \cos^2 x ] [ \sin^2 x = 1 - \left(\frac{3}{5}\right)^2 ] [ \sin^2 x = 1 - \frac{9}{25} ] [ \sin^2 x = \frac{25 - 9}{25} ] [ \sin^2 x = \frac{16}{25} ] [ \sin x = \pm \frac{4}{5} ]

Since (x) is in the first quadrant, (\sin x = \frac{4}{5}).

Now, use the double angle formulas:

[ \cos 2x = 2\left(\frac{3}{5}\right)^2 - 1 ] [ \cos 2x = 2\left(\frac{9}{25}\right) - 1 ] [ \cos 2x = \frac{18}{25} - 1 ] [ \cos 2x = \frac{18}{25} - \frac{25}{25} ] [ \cos 2x = -\frac{7}{25} ]

[ \sin 2x = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} ] [ \sin 2x = \frac{24}{25} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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