prove that : sin² pie/6 + cos²pie/3 - tan²pie/4 = - 1/4
Answers
Answer:
Taking LHS
sin ²pie /6+ cos²pie/3- tan²pie/4
Putting pie = 180'
sin²180/6+cos²180/3- tan²180/6
in²30'+cos²60'- tan²45'
(sin 30')²+(cos 60')²- (tan 45')²
Putting
sin30'=1/2
cos60'=1/2
tan45'=1
=(1/2)²+(1/2)²- (1)²
=1/4+1/4-1
=1+1-6/4
=2-4/4
= -2/4
= - 1/2
RHS
Hence proved.
Question : -
Prove that : sin² π/6 + cos² π/3 - tan² π/4 = -1/2
ANSWER
Given : -
sin² π/6 + cos² π/3 - tan² π/4 = -1/4
Required to prove : -
- LHS = RHS
Concept used : -
The standard values trigonometry angles are ;
Proof : -
sin² π/6 + cos² π/3 - tan² π/4 = -1/2
We need to prove that LHS = RHS .
So,
We know that ;
This implies ;
sin² 180°/6 + cos² 180°/3 - tan² 180°/4 = -1/2
sin² 30° + cos² 60° - tan 45° = -1/2
From the standard values of trigonometric angles we can say that ;
- sin 30° = 1/2
- cos 60° = 1/2
- tan 45° = 1
By substituting these values in the above one .
➞ ( 1/2 )² + ( 1/2 )² - (1 )² = -1/2
➞ 1/4 + 1/4 - 1 = -1/2
➞ 1 + 1/4 - 1 = -1/2
➞ 2/4 - 1 = -1/2
➞ 2/4 - 4/4 = -1/2
➞ - 2/4 = -1/2
➞ -1/2 = -1/2
➞ LHS = RHS
Hence Proved !
- : Additional Information : -
Some important formulae ;
sin ( A + B ) = sin A cos B + sin B cos A
sin ( A - B ) = sin A cos B - sin B cos A
cos ( A + B ) = cos A cos B - sin A sin B
cos ( A - B ) = cos A cos B + sin A sin B
tan ( A + B ) = tan A + tan B/1 - tan A tan B
tan ( A - B ) = tan A - tan B/1 + tan A tan B
___________________________________
sin @ = opposite side/Hypotenuse
cos @ = Adjacent side/Hypotenuse
tan @ = Opposite side/Adjacent side
cot @ = Adjacent side/Opposite side
sec @ = Hypotenuse/Adjacent side
cosec @ = Hypotenuse/Opposite side
Here, ( @ = theta )
___________________________
Complementary angles
sin @ = cos ( 90° - @ )
cos @ = sin ( 90° - @ )
tan @ = cot ( 90° - @ )
cot @ = tan ( 90° - @ )
sec @ = cosec ( 90° - @ )
cosec @ = sec ( 90° - @ )
Here , ( @ = theta )