Is this true or false? #sin^2(θ) ≠ sin(θ)^2#

Question

Don't understand this hw questions.. I believe it is false though?

14028 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-06T19:50:08

I would answer that it is true.

This hw question is a question about notation.

#sin^2(theta)# is the same as #(sin(theta))^2#

I would read #sin(theta)^2# as equivalent to #sin(theta^2)#

(But I've noticed that WolframAlpha treats #sin(theta)^2# as equal to #(sin(theta))^2#.

I would never write #sin(theta)^2# because it is ambiguous.

Answer 2 · 2018-04-06T19:51:37

True.

#sin^2(theta)!=sin(theta)^2#

If you read #sin(theta)^2# as #sin(theta^2)# then the statement is true.

#sin^2x# is the square of a ratio.

#sin(x^2)# is the square of an angle.

Answer 3 · 2018-04-06T20:09:21

True, #sin^2(theta) != sin(theta)^2#

Here's how I did it:

If you do #sin(x)^2#, that means you are squaring the angle, instead of the value of #sin(x)#. For example:

#sin(30^@)^2=sin(900^@) = sin(180^@) = 0#
is different from
#sin^2(30^@) = (1/2)^2 = 1/4#

Therefore, #sin^2(theta) != sin(theta)^2#.

The answer is true .

Hope this helps!