Trying to use this trig identity to find other angles with same ratios but ending up with wrong answer

B

bushra1175

Junior Member
Joined
Jun 14, 2020
Messages
59
I am using the identity cos θ = -cos(θ -180∘) as this is part of a list of identities in the textbook:
1598479161249.png


Also, the confusing thing is, this identity is also included so I'm not even sure I'm using the correct one:
1598479175388.png

Anyway, the question is below:

1598479236137.png

I substituted θ for 40∘ in cos θ = -cos(θ -180∘) which brings me to cos θ = -cos(40∘ -180∘), which is -cos(-140∘). This equals 0.7660. However, -cos 40∘ in the original expression equals -0.7660 so the two values don't equate each other. I feel like I'm using the identity wrong so I would appreciate the help.
 
Last edited:
bushra1175 said:
I am using the identity cos θ = -cos(θ -180∘) as this is part of a list of identities in the textbook:



Also, the confusing thing is, this identity is also included so I'm not even sure I'm using the correct one:
Click to expand...
What is the question?
 
Jomo said:
Note that (a-b) = -(b-a)
Also cos( -x ) = cos ( x ).

So cos( a - b ) = cos (-(a-b)) = cos (b-a)
Click to expand...

Thanks. I have now factored in that -cos 40o when plugged into the identity -cos(θ -180∘), has to be cos(θ -180∘) since - (-θo) = +θo. This makes cos(40∘ -180∘) which is cos(-140∘).

Since you pointed out that cos( a - b ) = cos (b-a) , this means cos( 180 - 40 ) = cos (40-180), and therefore θ = 140. However, the solution for this question below is also ending up with 220∘. How is this done?

1598521817265.png

I tried to use the other identity cos θ = cos (360-θ) and this is what I got:

cos θ = cos (360-θ)
cos( b - a ) = cos ( a - b )
cos( (-40) - 360 ) = cos ( 360 - (-40) )
cos( -400 ) = cos ( 400 )
 
cos θ = -cos(θ -180∘) is an identity. OK?

You want θ such that cos(θ) = - cos (40). Now this equation looks just like the equation above with θ-180 being replaced with 40. Do you see that?

So 40 = θ -180. Solving for θ yields θ = 220. Any questions?
 
Jomo said:
cos θ = -cos(θ -180∘) is an identity. OK?

You want θ such that cos(θ) = - cos (40). Now this equation looks just like the equation above with θ-180 being replaced with 40. Do you see that?

So 40 = θ -180. Solving for θ yields θ = 220. Any questions?
Click to expand...

According to your explanation, I have realised I was using identities wrong in my previous post. So I tried applying the expression cos θ = -cos 40o to the other identity cos θ = cos (360-θ) and see if I end up with 220 or 140 like in the textbook solution :

cos θ = cos (360-θ)
cos θ = cos (40)

40 = 360-θ
θ = 320

This result is not consistent with the ones in the solution. What am I doing wrong?

Also, since I didn't use the identity cos θ = -cos(θ -180∘)c correctly in my previous post, please can you show me how to end up with θ=140 since there are two answers and you only ended up with θ=220 ?

Sorry for all the questions, this is by far the hardest concept I've ever had to grasp in maths.
 
I asked three questions below which you failed to answer in my last post. I will give you one more chance to answer them. If you don't then I can't help you

cos θ = -cos(θ -180∘) is an identity. OK?

You want θ such that cos(θ) = - cos (40). Now this equation looks just like the equation above with θ-180 being replaced with 40. Do you see that?

So 40 = θ -180. Solving for θ yields θ = 220. Any questions?

Once we settle that 220 is an answer we will work on the 2nd solution.
 
Jomo said:
I asked three questions below which you failed to answer in my last post. I will give you one more chance to answer them. If you don't then I can't help you

cos θ = -cos(θ -180∘) is an identity. OK?

You want θ such that cos(θ) = - cos (40). Now this equation looks just like the equation above with θ-180 being replaced with 40. Do you see that?

So 40 = θ -180. Solving for θ yields θ = 220. Any questions?

Once we settle that 220 is an answer we will work on the 2nd solution.
Click to expand...

yes on all three questions. Sorry I thought they were rhetorical and you just wanted me to acknowledge your method
 
bushra1175 said:
Cos(θ) = -Cos(180-θ)
40 = 180-θ
40+θ = 180-θ+θ
40+θ = 180
θ = 180-40
θ = 140

Thanks
Click to expand...
You should always look for a way to check your answer.

140o is in 2nd quadrant - means cos(140o) < 0

This we claim is equal to [-cos(40o)].

40o is in first quadrant - so cos(40o) > 0. Hence [-cos(40o)] < 0 , just like cos(140o)

So the signs check. You may want to check the numerical values using a calculator.
 
Subhotosh Khan said:
You should always look for a way to check your answer.

140o is in 2nd quadrant - means cos(140o) < 0

This we claim is equal to [-cos(40o)].

40o is in first quadrant - so cos(40o) > 0. Hence [-cos(40o)] < 0 , just like cos(140o)

So the signs check. You may want to check the numerical values using a calculator.
Click to expand...

Hi Subhotosh. Yes I checked my answer on the calculator so that -cos(40o) = -0.7660, and so does cos(140o) and cos(220o)
 
You must log in or register to reply here.
Top