Trigonometry! #85

Geometry Level 3

If cos ⁡ θ + sin ⁡ θ = 2 cos ⁡ θ \cos \theta + \sin \theta = \sqrt{2} \cos \theta then the equivalent of cos ⁡ θ − sin ⁡ θ \cos \theta - \sin \theta is?

This problem is part of the set Trigonometry .

2 sin ⁡ θ \sqrt{2} \sin \theta 2 cos ⁡ θ + sin ⁡ θ \frac{\sqrt{2}}{\cos \theta + \sin \theta} 2 tan ⁡ θ \sqrt{2} \tan \theta 2 cos ⁡ θ \sqrt{2}\cos\theta

View wiki
Omkar Kulkarni by Omkar Kulkarni , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

6 solutions

From the first equation we have that sin ⁡ ( θ ) = ( 2 − 1 ) cos ⁡ ( θ ) . \sin(\theta) = (\sqrt{2} - 1)\cos(\theta).

So cos ⁡ ( θ ) − sin ⁡ ( θ ) = cos ⁡ ( θ ) − ( 2 − 1 ) cos ⁡ ( θ ) = \cos(\theta) - \sin(\theta) = \cos(\theta) - (\sqrt{2} - 1)\cos(\theta) =

( 2 − 2 ) cos ⁡ ( θ ) = 2 ( 2 − 1 ) cos ⁡ ( θ ) = 2 sin ⁡ ( θ ) . (2 - \sqrt{2})\cos(\theta) = \sqrt{2}(\sqrt{2} - 1)\cos(\theta) = \boxed{\sqrt{2}\sin(\theta)}.

11 Helpful 0 Interesting 0 Brilliant 0 Confused
Anandhu Raj
Feb 4, 2015

Given cos ⁡ θ + sin ⁡ θ = 2 cos ⁡ θ \cos \theta + \sin \theta = \sqrt{2} \cos \theta

We have ( c o s θ + s i n θ ) 2 + ( c o s θ − s i n θ ) 2 = 2 { (cos\theta +sin\theta ) }^{ 2 }+{ (cos\theta -sin\theta ) }^{ 2 }=2

⟹ ( 2 c o s θ ) 2 + ( c o s θ − s i n θ ) 2 = 2 \Longrightarrow { (\sqrt { 2 } cos\theta ) }^{ 2 }+{ (cos\theta -sin\theta ) }^{ 2 }=2

⟹ ( c o s θ − s i n θ ) 2 = 2 − 2 c o s 2 θ \Longrightarrow { (cos\theta -sin\theta ) }^{ 2 }=2-2{ cos }^{ 2 }\theta

⟹ ( c o s θ − s i n θ ) 2 = 2 ( 1 − c o s 2 θ ) \Longrightarrow { (cos\theta -sin\theta ) }^{ 2 }=2(1-{ cos }^{ 2 }\theta )

⟹ ( c o s θ − s i n θ ) 2 = 2 ( s i n 2 θ ) \Longrightarrow { (cos\theta -sin\theta ) }^{ 2 }=2({ sin }^{ 2 }\theta )

⟹ c o s θ − s i n θ = ± 2 s i n θ \Longrightarrow \boxed{{ cos\theta -sin\theta }=\pm \sqrt { 2 } { sin }\theta }

10 Helpful 0 Interesting 0 Brilliant 0 Confused
Titas Biswas
Jul 23, 2015

The answer may also be (minus) root over sin theta.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you explain yourselves?

Omkar Kulkarni - 5 years, 10 months ago

Log in to reply

Look at the solution above by Anandhu Raj.

Titas Biswas - 5 years, 10 months ago

Log in to reply

Look at the solution above by Brian Charlesworth.

Omkar Kulkarni - 5 years, 10 months ago

Log in to reply

@Omkar Kulkarni – Yes,I did go through that.But you cannot deny this answer,can you?

Titas Biswas - 5 years, 10 months ago

Log in to reply

@Titas Biswas – Agreed, but Brian's solution gives a more specified answer. Look at it this way: As both solutions must hold true, we must take the intersection of both the solution sets, which is the same as Brian's answer.

Omkar Kulkarni - 5 years, 10 months ago

0 Helpful 0 Interesting 0 Brilliant 0 Confused

\( \text {Squaring the given equation} \\ =>(sin(x))^2 + (cos(x))^2 + 2sin(x)cos(x) = 2(cos(x))^2 => 1 + sin(2x) = 2(cos(x))^2 => sin(2x) = cos(2x) -----(1) Let cos(x) - sin(x) = q

(cos(x) + sin(x))(cos(x) - sin(x)) = p(√2cos(x)) =>( cos(x))^2 - (sin(x))^2 = cos(2x) = sin(2x) = p(√2cos(x)) => 2sin(x) = p√2 => p = √2sin(x) \)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Lu Chee Ket
Feb 5, 2015

Sin q = [Sqrt (2) - 1] Cos q

Tan q = Sqrt (2) - 1

=> Sin q = [Sqrt (2) -1]/ Sqrt [4 - 2 Sqrt(2)] and Cos q = 1/ Sqrt [4 - 2 Sqrt(2)]

Cos q - Sin q = [2 - Sqrt (2)]/ Sqrt [4 - 2 Sqrt(2)]

Sqrt (2) Sin q

=Sqrt (2) [Sqrt (2) -1]/ Sqrt [4 - 2 Sqrt(2)]

= [2 - Sqrt (2)]/ Sqrt [4 - 2 Sqrt(2)]

Therefore, Sqrt (2) Sin q is the answer.

Note: Actually, I obtained answer by knowing q = 22.5 d

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...