# By considering the relationships between the sides of the right angled triangle (hypotonuese of 12 cm) explain why sin x can never be greater than 1?

QuestionSee Below.

#sin theta = "opposite"/"hypotenuse"#

That means, #"Opposite " lt " Hypotenuse"#.

So, #"Opposite"/"Hypotenuse" lt 1# [It is an proper fraction].

I.e #sin theta lt 1#.

We all know, #sin^2 theta + cos^2 theta = 1#

As, #cos^2 theta# is a square term, it to be real, has to be positive.

So, #cos^2 theta gt 0#

So, #-cos^2 theta lt 0#

#rArr 1 - cos^2 theta lt 1#

#rArr sin^2 theta lt 1# [From the identity]

#rArr sin^2 theta - 1 lt 0#

#rArr (sin theta + 1)(sin theta - 1) lt 0#

Now, Either #sin theta lt -1# or #sin theta lt 1#.

We know, Sine is a Periodic Function whose value ranges from #-1# to #1#. (#-1 <= sin x <= 1#).

So, #sin theta cancellt -1#. But, #sin theta lt 1#.

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If the value of the

# A^2 + B^2 = C^2 #

# sin = A # # cos = B# # hyp = C#

# sin^2 + cos^2 = hyp^2#

# sin^2 + cos^2 = 1^2 #

This can be illustrated by a #45# degree right triangle

#Sin = cos# so

#sin 45 = 0.707# #cos 45 = 0.707 #

# 0.707^2 + 0.707^2 =1^2 #

# 0.5 + 0.5 = 1 #

# 1 =1 # Classic Pythagorean theorem.

# 1^2 + 0^2 = 1^2#

# 1 = 1 #

So if the angles of the triangle are such that #sin =1 and cos =0 #there is no longer a triangle but just a vertical line because the line adjacent is 0

The value of the #Sin# cannot be more than 1 because the triangle that the trig functions are based on no longer exists.

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