If -2- then the minimum value of the expression 6sin-8sin-12 is

Given : nbsp;α+β=π2 sinβ=sin(π2 α)=cosα Now, 6sinα 8sinβ+12 =6sinα 8cosα+12 We know that, nbsp; 6sinα 8cosα∈ [ √(6^2+8^2), √(6^2+8^2)] ⇒ nbsp;6sinα 8co ...

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Standard XII

Mathematics

Range of Trigonometric Expressions

If α+β=π2, th...

Question

If $α + β = \frac{π}{2},$ then the minimum value of the expression $6 sin α - 8 sin β + 12$ is

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Solution

Given : $α + β = \frac{π}{2}$
$sin β = sin (\frac{π}{2} - α) = cos α$

Now,
$6 sin α - 8 sin β + 12 = 6 sin α - 8 cos α + 12$

We know that,
$6 sin α - 8 cos α \in [- \sqrt{6^{2} + 8^{2}}, \sqrt{6^{2} + 8^{2}}] \Rightarrow 6 sin α - 8 cos α \in [- 10, 10] \Rightarrow 6 sin α - 8 cos α + 12 \in [2, 22]$

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Range of Trigonometric Expressions

Standard XII Mathematics

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If α+β=π2, then the minimum value of the expression 6sinα−8sinβ+12 is

Given : α+β=π2 sinβ=sin(π2−α)=cosα Now, 6sinα−8sinβ+12=6sinα−8cosα+12 We know that, 6sinα−8cosα∈[−√62+82,√62+82]⇒6sinα−8cosα∈[−10,10]⇒6sinα−8cosα+12∈[2,22]

If $α + β = \frac{π}{2},$ then the minimum value of the expression $6 sin α - 8 sin β + 12$ is