A calculus problem by حسن العطية

Calculus Level 1

lim ⁡ θ → 0 ( cos ⁡ θ − sin ⁡ θ × cos ⁡ θ ) ( cos ⁡ θ + sin ⁡ θ × cos ⁡ θ ) = ? \lim_{\theta \to 0} {(\cos \theta -\sin \theta \times \cos \theta)(\cos \theta + \sin \theta \times \cos \theta)} = \ ?


The answer is 1.

حسن العطية by حسن العطية , 22, Saudi Arabia

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1 solution

lim ⁡ θ → 0 ( cos ⁡ θ − sin ⁡ θ × cos ⁡ θ ) ( cos ⁡ θ + sin ⁡ θ × cos ⁡ θ ) \displaystyle \lim _{\theta \to 0} \ \left( \cos { \theta } -\sin { \theta } \times \cos { \theta } \right) \left( \cos { \theta } +\sin { \theta } \times \cos { \theta } \right)

= lim ⁡ θ → 0 ( ( cos ⁡ θ ) 2 − ( sin ⁡ θ ) 2 × ( cos ⁡ θ ) 2 ) = \displaystyle \lim _{\theta \to 0} \left( (\cos \theta) ^{ 2 } -(\sin { \theta }) ^{ 2 }\times (\cos \theta) ^{ 2 } \right)

= lim ⁡ θ → 0 ( cos ⁡ θ ) 2 × ( 1 − ( sin ⁡ θ ) 2 ) = \displaystyle \lim _{\theta \to 0} \ ( \cos { \theta } )^{ 2 }\times \left( 1-(\sin { \theta }) ^{ 2 } \right)

= lim ⁡ θ → 0 ( cos ⁡ θ ) 2 × ( cos ⁡ θ ) 2 = \displaystyle \lim _{\theta \to 0} \ ( \cos { \theta } )^{ 2 }\times ( \cos { \theta } )^{ 2 }

= lim ⁡ θ → 0 ( cos ⁡ θ ) 4 = \displaystyle \lim _{\theta \to 0} \ (\cos { \theta }) ^{ 4 }

= ( cos ⁡ 0 ) 4 = (\cos { 0 }) ^{ 4 }

= 1 = \boxed{1}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice and simple problem involving trigonometric identities.

While writing powers of trigonometric functions, I would recommend that you make use of parentheses. For example, cos ⁡ θ 2 \cos \theta^{2} could be interpreted as cos ⁡ ( θ 2 ) \cos (\theta^2) which is incorrect. Instead, you could use ( cos ⁡ θ ) 2 (\cos \theta)^2 to avoid ambiguity.

To denote limits in Latex, it is easier to write them as \lim _{\theta \to 0} as compared to \underset{\theta \rightarrow 0} {lim} . It's easier to type and looks neat. I have edited your solution accordingly; you can have a look at it.

Pranshu Gaba - 5 years, 1 month ago

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Thank you so much

حسن العطية - 5 years, 1 month ago

Why cant you just simply plug in 0 at the start? You still end up getting 1.

Sarthak Sharma - 3 years, 6 months ago

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