For x = - 2 π
sin ( - 2 π ) = 0 , cos ( - 2 π ) = 1 , tan ( - 2 π ) = 0
sin ( - 2 π ) ∙ tan ( - 2 π ) = 0 ∙ 0 = 0 ≠ cos ( - 2 π )
0 ≠ 1
For x = - 3 π
sin ( - 3 π ) = 0 , cos ( - 3 π ) = - 1 , tan ( - 3 π ) = 0
sin ( - 3 π ) ∙ tan ( - 3 π ) = 0 ∙ 0 = 0 ≠ cos ( - 3 π )
0 ≠ - 1
For x = - 3 π / 4
sin ( - 3 π / 4 ) = -√ 2 / 2 , cos ( - 3 π / 4 ) = - √ 2 / 2 , tan ( - 3 π / 4 ) = 1
sin ( - 3 π / 4 ) ∙ tan ( - 3 π / 4 ) = -√ 2 / 2 ∙ 1 = - √ 2 / 2 = cos ( - 3 π / 4 )
- √ 2 / 2 = - √ 2 / 2
For x = - π / 4
sin ( - π / 4 ) = -√ 2 / 2 , cos ( - π / 4 ) = √ 2 / 2 , tan ( - π / 4 ) = - 1
sin ( - π / 4 ) ∙ tan ( - π / 4 ) = -√ 2 / 2 ∙ ( - 1 ) = √ 2 / 2 = cos ( - π / 4 )
√ 2 / 2 = √ 2 / 2
So x = - 2 π and x = - 3 π proves that
sin x ∙ tan x = cos x is not a trigonometric identity,
Which of the following counterexamples proves that sinxtanx=cosx is not a trigonometric identity? Select all that apply.
-2π
-3π
-3π/4
-π/4
4 answers
Seems like Reiny already helped you with this, but here goes again.
If it is an identity, then all of the values suggested should work.
So, does
sin(-2π)*tan(-2π) = cos(-2π) ?
0*0 = 1 ?
Nope. So, using -2π proves it is not an identity.
Now you try the others. If you get stuck, come on back and show what you got.
If it is an identity, then all of the values suggested should work.
So, does
sin(-2π)*tan(-2π) = cos(-2π) ?
0*0 = 1 ?
Nope. So, using -2π proves it is not an identity.
Now you try the others. If you get stuck, come on back and show what you got.
Also you can do this:
sin x ∙ tan x = cos x
sin x ∙ sin x / cos = cos x
sin² x / cos = cos x
Multiply both sides by cos x
sin² x = cos² x
For x = - 2 π
sin ( - 2 π ) = 0 , cos ( - 2 π ) = 1
sin² ( - 2 π ) = 0 ≠ cos² ( - 2 π )
0 ≠ 1²
0 ≠ 1
For x = - 3 π
sin ( - 3 π ) = 0 , cos ( - 3 π ) = - 1 ,
sin² ( - 3 π ) = 0 ≠ cos² ( - 3 π )
0 ≠ ( - 1 )²
0 ≠ 1
For x = - 3 π / 4
sin ( - 3 π / 4 ) = -√ 2 / 2 , cos ( - 3 π / 4 ) = - √ 2 / 2
sin² ( - 3 π / 4 ) = ( -√ 2 / 2 )² = 2 / 4 = 1 / 2 =
cos² ( - 3 π / 4 ) = ( - √ 2 / 2 )² = 2 / 4 = 1 / 2
1 / 2 = 1 / 2
For x = - π / 4
sin ( - π / 4 ) = -√ 2 / 2 , cos ( - π / 4 ) = √ 2 / 2
sin² ( - π / 4 ) = ( -√ 2 / 2 )² = 2 / 4 = 1 / 2 =
cos² ( - π / 4 ) = ( - √ 2 / 2 )² = 2 / 4 = 1 / 2
1 / 2 = 1 / 2
So x = - 2 π and x = - 3 π proves that
sin x ∙ tan x = cos x is not a trigonometric identity.
sin x ∙ tan x = cos x
sin x ∙ sin x / cos = cos x
sin² x / cos = cos x
Multiply both sides by cos x
sin² x = cos² x
For x = - 2 π
sin ( - 2 π ) = 0 , cos ( - 2 π ) = 1
sin² ( - 2 π ) = 0 ≠ cos² ( - 2 π )
0 ≠ 1²
0 ≠ 1
For x = - 3 π
sin ( - 3 π ) = 0 , cos ( - 3 π ) = - 1 ,
sin² ( - 3 π ) = 0 ≠ cos² ( - 3 π )
0 ≠ ( - 1 )²
0 ≠ 1
For x = - 3 π / 4
sin ( - 3 π / 4 ) = -√ 2 / 2 , cos ( - 3 π / 4 ) = - √ 2 / 2
sin² ( - 3 π / 4 ) = ( -√ 2 / 2 )² = 2 / 4 = 1 / 2 =
cos² ( - 3 π / 4 ) = ( - √ 2 / 2 )² = 2 / 4 = 1 / 2
1 / 2 = 1 / 2
For x = - π / 4
sin ( - π / 4 ) = -√ 2 / 2 , cos ( - π / 4 ) = √ 2 / 2
sin² ( - π / 4 ) = ( -√ 2 / 2 )² = 2 / 4 = 1 / 2 =
cos² ( - π / 4 ) = ( - √ 2 / 2 )² = 2 / 4 = 1 / 2
1 / 2 = 1 / 2
So x = - 2 π and x = - 3 π proves that
sin x ∙ tan x = cos x is not a trigonometric identity.
Or, having arrived at
sin² x = cos² x
then
tan² x = 1
which is only true for odd multiples of π/4
so, for -2π and -3π it won't work
sin² x = cos² x
then
tan² x = 1
which is only true for odd multiples of π/4
so, for -2π and -3π it won't work