If (fθ)= (1-sin2θ + cos2θ)/2cos2θ then Find about F(11^0) f((34^0) ?
In the realm of mathematics, functions often take center stage as they help us model and understand various phenomena. One such function is
f(θ)=2cos2(2θ)1−sin2(2θ)+cos2(2θ), which involves trigonometric expressions and can be quite intricate to evaluate. In this discussion, we will delve into the intricacies of this function and compute its values for specific angles, namely 11∘and 34∘. Through this exploration, we aim to unravel the beauty of mathematical functions and showcase the practicality of trigonometric concepts in solving real-world problems.
Solution:
(fθ)= (1-sin2θ + cos2θ)/2cos2θ
sin2θ= 2sinθcosθ
cos2θ=cos2 θ −sin2θ
(fθ)=(1- sin2θ + cos2θ)/2 cos2 θ
(fθ)=( cos2 θ + sin2θ- 2sinθcos2θ) +( cos2 θ −sin2θ)/ 2(cos2 θ −sin2θ)
(fθ)=( cos θ – sinθ)^2 +( cos θ + sinθ)( )( cos θ – sinθ)/ )( cos θ – sinθ)( )( cos θ + sinθ)
Now )=( cos θ – sinθ is a comman factor then
(fθ)=( cos θ – sinθ) )( cos θ + sinθ)+( cos θ – sinθ)/2(cos θ – sinθ )(cos θ + sinθ)
Now cancel the (cos θ – sinθ) with (cos θ – sinθ)
(fθ)=( cos θ + sinθ)+( cos θ – sinθ)/2)(cos θ + sinθ)
For complete the solution, watch the above video:
Conclusion
In conclusion, the evaluation of f(θ) at specific angles not only enriches our understanding of mathematical functions but also reaffirms the timeless importance of mathematics as a cornerstone of human knowledge and innovation. As we continue to explore the boundless realms of mathematics, we discover that its beauty and utility are, indeed, infinite.
