How do you find the exact value of #cos^-1 (sqrt2/2)#?

Answer 1

See below

Let #theta = cos^-1(sqrt2/2)#

#costheta=sqrt2/2#

So in an imaginary right-angled triangle, the length of the #"hyp"# is #2# and the length of the #"adj"# is #sqrt2#. This means that the length of the #"opp"# is #sqrt(2^2-(sqrt2)^2)=sqrt2#.

Since the #"adj"# and #"opp"# are equal lengths, our triangle is isosceles. This means that it also has two angles of equal lengths. Since one angle is #pi/2#, the other two must be #pi/4#.

So if we say that #theta=pi/4#, then #cos(pi/4)="adj"/"hyp"=sqrt2/2# #thereforepi/4=cos^-1(sqrt2/2)#

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Answer 2

Carter Anderson

To find the exact value of $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)$ , you need to determine the angle whose cosine equals $\frac{\sqrt{2}}{2}$ .

Since $\frac{\sqrt{2}}{2}$ is the cosine of $\frac{\pi}{4}$ radians (or $45^\circ$ ), the exact value of $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)$ is $\frac{\pi}{4}$ radians (or $45^\circ$ ).

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Answer from HIX Tutor

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.