Factor Trigonometric Identities

Topics

Try VividMath Premium to unlock full access

Start Free Trial

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

Question 1 of 2

1. Question
Factorise

$2 {sec}^{2} A - 3 sec A + 1$ $2sec^2A-3secA+1$
- 1. $(2 sec - 1) (sec - 1)$ $(2\text(sec)-1)(\text(sec)-1)$
- 2. $(2 sec + 1) (sec + 1)$ $(2\text(sec)+1)(\text(sec)+1)$
- 3. $(2 sec + 1) (sec - 1)$ $(2\text(sec)+1)(\text(sec)-1)$
- 4. $(2 sec + 1) (2 + sec)$ $(2\text(sec)+1)(2+\text(sec))$
Hint

Help Video
Correct

Excellent!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

The cross method is a factorisation method used for quadratics.

Since the equation is in standard form $($ $($ $a$ $a$ $x^{2} +$ $x^2+$ $b$ $b$ $x +$ $x+$ $c$ $c$ $= 0)$ $=0)$ we can factorise using the cross method.

$2 {sec}^{2} A$ $2\text(sec)^2A$ $- 3 sec A$ $-3\text(sec)A$ $+ 1$ $+1$ $= 0$ $=0$

To factorise, we need to find two values on the left side that multiply to $2 {sec}^{2} A$ $2\text(sec)^2A$ and two values on the right side that multiply to $1$ $1$ and, when cross multiplied with the left side values and added together, gives $- 3 sec A$ $-3\text(sec)A$

For the left side, $2 sec A$ $2\text(sec)A$ and $sec A$ $\text(sec)A$ fit the condition

$2 sec A \times sec A$ $2\text(sec)A xx \text(sec)A$ $=$ $=$ $2 {sec}^{2} A$ $2\text(sec)^2A$

For the right side, $- 1$ $-1$ and $- 1$ $-1$ fit both conditions

$[2 sec A \times (- 1)] + [sec A \times (- 1)]$ $[2\text(sec)Axx(-1)]+[\text(sec)Axx(-1)]$ $=$ $=$ $- 3 sec A$ $-3\text(sec)A$

$- 1 \times - 1$ $-1 xx -1$ $=$ $=$ $1$ $1$

Read across to get the factors.

$(2 sec - 1) (sec - 1)$ $(2\text(sec)-1)(\text(sec)-1)$

$(2 sec - 1) (sec - 1)$ $(2\text(sec)-1)(\text(sec)-1)$

Question 2 of 2

Factorise

2 {sin}^{2} θ + 7 sin θ cos θ + 3 {cos}^{2} θ

$2sin^2 theta+7sin theta cos theta +3cos^2 theta$

1. $(sin θ - 2 cos θ) (3 sin θ - cos θ)$ $(\text(sin) theta-2 \text(cos) theta)(3 \text(sin) theta-\text(cos) theta)$
2. $(2 sin θ + cos θ) (sin θ + 3 cos θ)$ $(2 \text(sin) theta+\text(cos) theta)(\text(sin) theta+3 \text(cos) theta)$
3. $(3 sin θ + cos θ) (sin θ + 2 cos θ)$ $(3 \text(sin) theta+\text(cos) theta)(\text(sin) theta+2 \text(cos) theta)$
4. $(2 sin θ + 3 cos θ) (2 sin θ + 3 cos θ)$ $(2 \text(sin) theta+3 \text(cos) theta)(2 \text(sin) theta+3 \text(cos) theta)$

$(cos θ \times sin θ) + (3 cos θ \times 2 sin θ)$ $(\text(cos) thetaxx\text(sin) theta)+(3 \text(cos) thetaxx2 \text(sin) theta)$	$=$ $=$	$7 sin θ cos θ$ $7 \text(sin) theta \text(cos) theta$
$cos θ \times 3 cos θ$ $\text(cos) theta xx 3 \text(cos) theta$	$=$ $=$	$3 {cos}^{2} θ$ $3 \text(cos)^2theta$

Factor Trigonometric Identities

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Excellent!

2. Question

Hint

Well Done!

Quizzes

$[2 sec A \times (- 1)] + [sec A \times (- 1)]$ $[2\text(sec)Axx(-1)]+[\text(sec)Axx(-1)]$	$=$ $=$	$- 3 sec A$ $-3\text(sec)A$
$- 1 \times - 1$ $-1 xx -1$	$=$ $=$	$1$ $1$