at [tex]\cos \emptyset-\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)=0[/tex] ii. [tex]\tan 15^{\circ}=2-\sqrt{3}[/tex]
Answer (2)
The equation simplifies to cos ∅ = 26 , which is beyond the valid range of the cosine function. Therefore, there is no real solution for ∅ .
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Rewrite the given equation: cos ∅ = 2 1 ( a 3 + a 3 1 ) .
Substitute a = tan 1 5 ∘ = 2 − 3 into the equation.
Calculate ( 2 − 3 ) 3 = 26 − 15 3 and ( 2 − 3 ) 3 1 = 26 + 15 3 .
Substitute these values back into the equation to get cos ∅ = 26 .
Since cos ∅ cannot be 26, there is no real solution for ∅ . No real solution
Explanation
Problem Setup We are given the equation cos ∅ − 2 1 ( a 3 + a 3 1 ) = 0 and the value tan 1 5 ∘ = 2 − 3 . Our goal is to find the value of ∅ .
Rewriting the Equation First, let's rewrite the given equation as cos ∅ = 2 1 ( a 3 + a 3 1 ) . We are also given that tan 1 5 ∘ = 2 − 3 . Let's assume that a = tan 1 5 ∘ = 2 − 3 .
Substitution Now, substitute a = 2 − 3 into the equation cos ∅ = 2 1 ( a 3 + a 3 1 ) . We have cos ∅ = 2 1 (( 2 − 3 ) 3 + ( 2 − 3 ) 3 1 ) .
Calculating (2 - sqrt(3))^3 Let's calculate ( 2 − 3 ) 3 . Using the binomial expansion, we get ( 2 − 3 ) 3 = 2 3 − 3 ( 2 2 ) ( 3 ) + 3 ( 2 ) ( 3 ) 2 − ( 3 ) 3 = 8 − 12 3 + 18 − 3 3 = 26 − 15 3 .
Calculating 1/(2 - sqrt(3))^3 Now, let's calculate ( 2 − 3 ) 3 1 . Since ( 2 − 3 ) 3 = 26 − 15 3 , we have ( 2 − 3 ) 3 1 = 26 − 15 3 1 . To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is 26 + 15 3 . So, 26 − 15 3 1 = ( 26 − 15 3 ) ( 26 + 15 3 ) 26 + 15 3 = 2 6 2 − ( 15 3 ) 2 26 + 15 3 = 676 − 675 26 + 15 3 = 26 + 15 3 .
Substituting Back Now, substitute these values back into the equation cos ∅ = 2 1 (( 2 − 3 ) 3 + ( 2 − 3 ) 3 1 ) . We have cos ∅ = 2 1 (( 26 − 15 3 ) + ( 26 + 15 3 )) = 2 1 ( 52 ) = 26 .
Final Answer Since the range of the cosine function is [ − 1 , 1 ] , and we found that cos ∅ = 26 , there is no real solution for ∅ .
Examples
Understanding trigonometric equations is crucial in fields like physics and engineering. For instance, determining the angle at which a projectile is launched to achieve maximum range involves solving trigonometric equations. Similarly, in electrical engineering, analyzing alternating current circuits requires solving equations involving trigonometric functions to understand the phase relationships between voltage and current. These applications highlight the practical importance of mastering trigonometric concepts.
