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Popular >

-sin(θ)+cos(θ)-sqrt(10)<0

Solution

- sin (θ) + cos (θ) - 10 < 0

Solution

True for all θ \in R

Interval Notation

(- \infty, \infty)

Solution steps

- sin (θ) + cos (θ) - 10 < 0

Use the following identity:

cos (x) - sin (x) = 2 cos (\frac{π}{4} + x)

- 10 + 2 cos (\frac{π}{4} + θ) < 0

Move

10

to the right side

- 10 + 2 cos (\frac{π}{4} + θ) < 0

Add

10

to both sides

- 10 + 2 cos (\frac{π}{4} + θ) + 10 < 0 + 10

Simplify

2 cos (\frac{π}{4} + θ) < 10

2 cos (\frac{π}{4} + θ) < 10

Divide both sides by

2

2 cos (\frac{π}{4} + θ) < 10

Divide both sides by

2

\frac{2 cos ( \frac{π}{4} + θ )}{2} < \frac{10}{2}

Simplify

\frac{2 cos ( \frac{π}{4} + θ )}{2} < \frac{10}{2}

Simplify

\frac{2 cos ( \frac{π}{4} + θ )}{2} : cos (\frac{π}{4} + θ)

\frac{2 cos ( \frac{π}{4} + θ )}{2}

Cancel the common factor:

2

= cos (\frac{π}{4} + θ)

Simplify

\frac{10}{2} : 5

\frac{10}{2}

Factor

10 : 25

Factor

10 = 2 \cdot 5

= 2 \cdot 5

Apply radical rule:

= 25

= \frac{2 5}{2}

Cancel the common factor:

2

= 5

cos (\frac{π}{4} + θ) < 5

cos (\frac{π}{4} + θ) < 5

cos (\frac{π}{4} + θ) < 5

Range of

cos (\frac{π}{4} + θ) : - 1 \leq cos (\frac{π}{4} + θ) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (\frac{π}{4} + θ) \leq 1

- 1 \leq cos (\frac{π}{4} + θ) \leq 1

cos (\frac{π}{4} + θ) < 5 and - 1 \leq cos (\frac{π}{4} + θ) \leq 1 : - 1 \leq cos (\frac{π}{4} + θ) \leq 1

Let

y = cos (\frac{π}{4} + θ)

Combine the intervals

y < 5 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < 5 and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y < 5

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all θ

True for all θ \in R

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