Given that sin x = \(\frac{3}{5}\), 0 ≤ x ≤ 90, evaluate (tan x + 2cosx)

Given that sin x = \(\frac{3}{5}\), 0 ≤ x ≤ 90, evaluate (tan x + 2cosx)

Sin x = \(\frac{opp}{hyp}\)

sinx = \(\frac{3}{5}\)

using Pythagoras' theorem

hyp\(^2\) = opp\(^2\) + adj\(^2\)

adj\(^2\) = 5\(^2\) - 3\(^2\) = 25 - 9

adj\(^2\) = 16

adj = √ 16

adj = 4.

tanx = \(\frac{opp}{adj}\)

= \(\frac{3}{4}\)

cosx = \(\frac{adj}{hyp}\)

= \(\frac{4}{5}\)

(tanx + 2cosx) = \(\frac{3}{4}\) + 2(\(\frac{4}{5}\))

= \(\frac{15 + 32}{20}\)

= \(\frac{47}{20}\) or

2 \(\frac{7}{20}\)

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