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Answer: The value of sin135° is $\dfrac{1}{\sqrt 2}$ and the value of sin(-135°) is -$\dfrac{1}{\sqrt 2}$
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| Evaluate sin135° and sin(-135°) without using calculator. |
Evaluating sin135° using the relation of
sin(90° +A) = cos A.
sin(90°+A) = cosA is an identity to find the values of sin when A is an angle less than 90°.
In our question, we have 135°. Let us divide this angle into (90°+A).
We have,
135° = 90°+A
or, A = 135°-90°
So, A = 45°
Now, that we have the value of A= 45°, we can now find the value of sin135°.
We know,
sin135° = cos(90°+45°)
or, sin135° = cos45°
So, sin135° = $\dfrac{1}{\sqrt 2}$
Evaluating sin135° using the relation of sin(180°-A) = sin A.
sin(180°-A) = sinA is an identity to find the values of sin when A is an angle less than 90°.
In our question, we have 135°. Let us divide this info (180°-A).
We have,
135° = 180°-A
or, A = 180°-135°
So, A = 45°
Now, that we have the value of A = 45°, we can now find the value of sin135°.
We know,
sin135° = sin(180°-45°)
or, sin135° = sin45°
So, sin135° = $\dfrac{1}{\sqrt 2}$
Evaluating sin135° using the formula of sin(A+B).
We know,
sin(A+B) = sinA.cosB + cosA.sinB
In this formula, we need to divide the given compound angles into such angles whose values are known from the value table of trigonometric ratios. Like: 0°, 30°, 45°, 60° or 90°.
We can divide 135° into 90° and 45°.
Let A = 90° and B = 45°
sin135° = sin(90°+45°)
or, sin135° = sin90°.cos45° + cos90°.sin45°
or, sin135° = 1. $\dfrac{1}{\sqrt 2}$ + 0. $\dfrac{1}{\sqrt 2}$
So, sin135° = $\dfrac{1}{\sqrt 2}$
Evaluating sin135° using the formula of sin(A-B).
We know,
sin(A-B) = sinA.cosB - cosA.sinB
According to this formula, we can divide 135° into (180°-45°).
Let A = 180° and B = 45°
sin135° = sin(180°-45°)
or, sin135° = sin180°.cos45° - cos180°.sin45°
or, sin135° = sin(90°+90°). $\dfrac{1}{\sqrt 2}$ -cos(90°+90°).$ \dfrac{1}{\sqrt 2}$
or, sin135° = cos90°.$ \dfrac{1}{\sqrt 2}$ - (-sin90°).$ \dfrac{1}{\sqrt 2}$
or, sin135° = 0.$ \dfrac{1}{\sqrt 2}$ - (-1).$ \dfrac{1}{\sqrt 2}$
So, sin135° = $ \dfrac{1}{\sqrt 2}$
Evaluating sin(-135°) using sin(-90°-A) = -cosA.
We know, A = 45°.
sin135° = sin(-90°-45°)
or, sin135° = -cos45°
So, sin135° = -$ \dfrac{1}{\sqrt 2}$
Evaluating sin(-135°) using sin(A-B).
Let A = 45° and B = 180°
sin(45°-180°) = sin45°.cos180° -cos45°.sin180°
or, sin(-135°) = $ \dfrac{1}{\sqrt 2}$.(-1) - $ \dfrac{1}{\sqrt 2}$.0
So, sin(-135°) = - $ \dfrac{1}{\sqrt 2}$
Evaluating sin(-135°) using sin(A+B).
Let A = -90° and B = -45°
sin{-90° +(-45°)} = sin(-90°).cos(-45°) +cos(-90°).sin(-45°)
or, sin(-135°) = (-1)$ \dfrac{1}{\sqrt 2}$ + 0.(-$ \dfrac{1}{\sqrt 2}$
So, sin(-135°) = -$ \dfrac{1}{\sqrt 2}$
So, these were some of the many ways to find the values of sin135° and sin(-135°).
You can get more help by visiting these useful links:
Link: Compound Angles
Link: Values of Trigonometric Ratios
Link: Introduction To Trigonometry
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